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# CALCULUS Math 2B

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Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 8 Notes These notes correspond to Section 62 in the text Volume by Slices Suppose that S is a solid that lies bet 39 en the planes J a and J I If AJ represents the area of the crosssection or slice obtained by intersecting S with the plane that crosses the J axis at J and is perpendicular to the J axis then the volume of S is b V AJ dJ 1 l a This formula is a generalization of the formula for the volume of a cylinder of radius 9quot and height b I placed parallel to the J axis which is V m 2b a In this case the solid has a constant radius independent of J and therefore each cross section has constant area AJ E m Z The more general formula 1 can be obtained by dividing 0 b into subintervals J0J1 Jn1Jn where Jo a and Jn I On each subinterval Ji1Ji we then approximate the portion of S between the planes Ji1 and J 75 by a cylinder of height 75 Ji1 and radius AJi Just the definite integral allows us to compute the area of a region with nonconstant height it allows us to compute the volume of a solid whose cross sectional area is non constant Some solids whose volumes can be computed using the formula 1 are known solids of revolution they can be obtained by revolving the area of some twodimensional region around a line For such solids the cross sectional area AJ is typically very to determine To illustrate we discuss some examples 0 Suppose that the region in question is the area below the curve 9 fJ where fJ 2 0 from J a to J 1 Furthermore suppose that this region is revolved around the J axis to obtain the solid S Then for each J in at the cross section at J is a disc of radius fJ which implies AJ 7rf72 The volume can then be computed by using equation 1 with this choice of AJ This approach to computing the volume of such a solid is called the disc method Suppose that the region is the area between two curves 9 fJ and y gJ where fJ 2 gJ 2 0 from J a to J I As before the region is revolved around the J is Then for each J betw en a and l the cros ection of the solid at J is the region between V The function fJ determines the outer radius of the cross section and gJ determines the irmer rddius It follows that AJ 7rfJ2 gJ2 that is AJ is the difference of the areas of two concentric circles Because each cross ction resembles a washer the method of computing the volume of such a solid by integrating AJ is called the washer method Many solids howev are not solids of revolution In such ca the cross s ctional area AJ may still be e r to determine For example the crosssection may be a triangle or a rectangle If the cros section is not a shape whose area is to compute for each J then it may be necessary to intersect the solid with planes that are perpendicular to the t axis or the zaxis and determine the cross sectional area a function of y or 2 instead of a function of It is important to recognize the parallel between computing areas and computing volumes using definite integrals In both cases an object is approximated by n simpler objects such rectangles or cvlinders so that the desired quantity be obtained by computing it easily for each of the simpler objec s and adding the results to obtain an approximation R which is a Riemann sum Then by determining the overall result a function of n and computing linlna00 Rn the exact result is obtained It is a interesting exercise to inquire to what other simple formulas can be generalized to more dif cult problems using this approach Example 1 We will compute the volume of conical beam a cross section of which is shown in Figure 1 For each J between 0 and r the cross section of the beam at J can be viewed a rectangle of length and height 2J with a halfdisc of radius J removed from each side It follows that the cross sectional area is given by AJ 672 7rJ2 Using the formula in equation 1 we can compute volume of this solid follow 5 6J2 7rJ2dJ2J3 7 1252 3 2 l0 3 C Example 2 Suppose that we form a solid by revolving the region under 9 J2 from J 0 to J 1 around the The resulting solid is a curved funnel with crosr rectional area AJ 7FJ3922 It follows that its volume is 1 7amp39J4177T 0 lt Cl Example 3 Suppose that we form a solid by revolving the same region in Example 2 except that in this case instead of revolving it around the Jaxis we revolve it around line 1 2 Then the cross section of J is a disc with radius J2 2 instead of J2 It follows that the volume of this solid is 1 7 lt 0 4 83 3 4 T 4 5 3 1 1 J 41J393 7T 222d7f4424d7f 4J 0 i0 Cl Crosssection of conical beam Figure 1 Cross section of conical beam for some J between 0 and 5 Example 4 Suppose that we forni a solid by revolving the region between 9 J and y J2 around Then for each J between 0 and 1 where the curves 9 J and y J2 intersect the cross section of the solid at J is a washer with outer radius J and inner radius J on 0 1 It follows that the volume of the solid obtained by the washer niethod is 1 a 3 1 1 1 2 7rJ2 7rJ4dJ7rJ 7 439 0 J a 0 J a 1amp3 Example 5 Suppose that we forni a solid by revolving the region under 9 2 from J 0 to J 1 around line 9 2 This region is illustrated in Figure 2 For each J between 0 and 1 the cross ction of the solid at J is a washer The outer radius of the washer is 2 the distance between the line 9 2 and the line 9 0 The inner radius is the distance between the curve 9 2 and the line 9 2 which is 2 2 It follows that the volume of the solid is since J 2 L41 V C 25 2 2 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 15 7 y2X12 gt 1 i i 05 v v G o5 l 05 0 05 1 1 5 x Figure 2 The region bounded by y 2 9 0 J 0 and J 1 is to be revolved around the line 9 2 1 A 1 2 l F22 2d77 4 dJ7r4J 0 l 0 Cl Example 6 Suppose that we form a solid of revolution using the same region in Example 5 shown in Figure 2 except that this tinle we revolve the region around the taxis which is the vertical line J 0 In this case we can still use the formula in equation 1 except that the cross sectional area is now a function of 9 For each 9 between 0 and 2 the cross section at y is a disc For 0 g y g 1 the radius of the disc is equal to 1 and for 1 g y g 2 the radius is equal to the horizontal distance between the line J 0 and the curve 9 2 Rewriting the equation of the curve a function of y we obtain J 2 y2 and it follows that the radius of the disc is 2 ygt2for1ys2 Because two functions are needed to describe the cross sectional area 149 one for 0 g y g 1 and one for 1 g y g 2 it is necessary to break up the integral of 149 from 0 to 2 into two integrals Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 25 Notes These notes correspond to Section 91 in the text Arc Length In this lecture we will learn how to use calculus to compute the length of a curve that is described by an equation of the form J for some given function Just we learned how to compute the area under such a curve the limit of a sum of areas of simpler regions namely rectangles we can compute the length of the curve by interpreting the length a limit of a sum of lengths of the simplest curves known which are line segments Suppose that we wish to compute the length of the curve 1 f 1 between 1 a and 1 b We can approximate this length by dividing the interval 0 1 into subintervals of length A1 11 0 n just we did when we were trying to compute the approximate area under 9 f 1 Consider any subinterval 1541 Then if A1 is chosen to be suf ciently small the length of the curve 9 f 1 between 1 11 and 1 1 can be well approximated by the length of the line segment between the points 151f 174 and 1 f17 This line segment is the hypotenuse of a right triangle with legs of length A1 and f1i f17 1 and therefore the length L of the curve 9 f 1 between 1 151 and 1 1 is approximately f 1 f A 2 WP 2 L1 A1 f11 f1771 A1 A fwd 37 2 1 7JH gt A1 2 LNiLrn 1 MZAJ E N 7 A1 l l 7 As 7 the number of line segments approaches 00 A1 approaches zero so the length of each subinterval 1i11i tends to zero It follows from the definition of the derivative that I fUH fUH il f 739i71 A1 f0 im lim Axgt0 1 131 Mao A1 f3917gt1 3 and therefore the sum converges to the definite integral L grain b 1 f 1lzd1z 4 i1 a The value of this integral is called the arc length of the curve 9 fJ from J a to J b Similarl if a curve is defined by the equation J f y from y c to t d the arc length of the curve is given by the definite integral L41 V d J1 way z Example 1 Compute the arc length of the curve 9 2J 3 Where 0 3 J g 2 Solution Since the curve is just a line segment we can simply use the distance formula to compute the arc length since the arc length is the distance between the endpoints of the segment The endpoints are 0 3 and 2 7 and therefore the arc length is 2 027 32 2242202 6 Using the arc length formula we have 9 2 and therefore the arc length is given by the integral 2122d 25d 2 0 0 l V 2 0 Cl Example 2 Compute the arc length of the curve 9 sinJ from J 0 to J 7r Solution Since 9 cos J the arc length is given by the integral 1 cos2JdJ 8 i 0 Unfortunately this integral cannot be evaluated using the Fundamental Theorem of Calculus Using an approximation method such Simpsonquots Rule the value of the integral is seen to be approximately 38202 E Example 3 Compute the arc length of the astroid described by the equation J 3 923 1 Solution We consider only the portion of the astroid in the upper quadrant J 2 0 y 2 0 which has endpoints 01 and 10 In this quadrant the astroid can be described by the equation 9 1 x2332 9 It follows that the arc length L of this segment of the astroid is given by the integral 1 L 1y 2dJ i0 Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 23 Notes These notes correspond to Section 87 in the text Approximate Integration The Fundamental Theorem of Calculus provides a simple method for evaluating many definite integrals of the form t rain 1 r the theorem can only be used under the following conditions Howe o The integrand f 1 must be a known continuous function This is not always the case since knowledge of f 1 may be limited to a small set of values at specific sampling points in the interval 0 b where such values are typically obtained from measurements 0 Even if a formula for is known we must still be able to anti differentiate If no antiderivative is known then we cannot apply the Fundamental Theorem of Calculus In real applications integrals for which the Fundamental Theorem of Calculus is of no use arise quite frequently Such integrals cannot be evaluated exactl 39 and therefore we must approximate their values instead This leads to the study of numerical integration which is also known numerical quadrature Riemann Sums Since the definite integral is defined to be the limit of a sequence of Riemann sums certainly any valid Riemann sum can be used to approximate the value of the integral we have already used a few particular types of Riemann sums to obtain such an approximation 0 We can divide the interval 00 into n subintervals of equal width A1 l an and approximate the area under the graph of f by adding the areas of n rectangles of width A1 and height determined by the value of f 1 at the left endpoint of each subinterval The resulting Riemann sum is lib fJ 1a a it fJi1AJ xi a iAx 2 1 i1 0 Similarly we can determine the height of each rectangle using the right endpoint of each subinterval resulting in the approximation 4 n now a 2 may 3 i1 0 We have seen that a more accurate approximation can be obtained by setting the height of each rectangle equal to the value of f 1 at the midpoint of the corresponding subinterval The resulting approximation is known the Midpoint Rule 14 in b n ow a minam new f 4 1 i1 In addition 7 can approximate the integral by averaging the approximations obtained by using left endpoints and right endpoints The resulting rule called the Trapezoidal Rule effectively approximates the area under the graph of by n trapezoids of width A1 and heights 151 and ag for i 1n Intuitively it makes sense that an approximation using trapezoids is more accurate than using rectangles This is true in that the Trapezoidal Rule is far more accurate than using left or right endpoints In fact not only is it more accurate but the approximation obtained using the Trapezoidal Rule converges to the exact value of the integral at a much faster rate than the approximation obtained using left or right endpoints Specifically the error in the approximation by left or right endpoints decreases by a factor of 2 when the number of rectangles is doubled In other words the error is proportional to Ax and therefore we that these approximations converge linearly to the exact value n the number of rectangles tends to 00 The error in the Trapezoidal Rule however decreases by a factor of 4 when the number of rectangles is doubled It follows that the error is proportional to AK and therefore we that the approximation obtained using the Trapezoidal Rule converges quadratically to the exact value n the number of trapezoids tends to 00 Surprisingly it turns out that the Midpoint Rule even though it approximates the area using rectangles is even more accurate than the Trapezoidal Rule Its approximations converge quadrat ically to the exact value and the error with n rectangles is approximately half of the error of the Trapezoidal Rule using n trapezoids The reason for this greater accuracy can be seen by examining a particular subinterval 7145 of 11 Let 1 be the midpoint of this subinterval Then the area of the rectangle of height f 11 and width A1 is equal to the area of the trapezoid whose upper side is tangent to the graph of f at 1 since such a trapezoid bv the equation of the tangent line would have heights fJM f JMAJ2 and f 1m f JAt1AJ2 The fact that the upper side of this trapezoid is tangent to the graph of f allows this side to more closely approximate the graph of on this subinterval than the line passing through 151 and which is the upper side of the trapezoid used by the Trapezoidal Rule Simpson s Rule and Other Approximation Methods The left and right endpoint approximations to the definite integral approximate the integrand by a constant function on each subinterval of 11 to determine the height of each rectangle The Midpoint Rule and Trapezoidal Rule approximate by a linear function on each subinterval which helps explain why they are more accurate It is natural to ask if even more accuracy can be obtained by approximating by a higherdegree polynomial on each subinterval The answer to this question is s and this fact is the basis of Simpson s Rule This rule approximates f 1 by a quadratic function on each consecutive pair of subintervals where the number of subintervals is assumed to be even For each pair of subintervals 445 and 13 741 Simpson s Rule implicitly constructs the unique quadratic function that past s through the three endpoints 721 3 and n1 This quadratic function is then integrated exactly on the interval 11J1 using the Fundamental Theorem of Calculus This integral turns out to be equal to A1 3 fUH il WU fJ39i1l 5 Where A1 I an is the width of each subinterval Adding the results from all pairs of subintervals we obtain the approximation 4 W a a A l a This rule exhibits fourthorder convergence that is doubling the number of subintervals reduces the error by a factor of 16 All of the rules we have discussed are examples of tnterpolatory quadrature rules These rules have the property that they approximate the integrand by polynomials on subintervals of 11 and then integrate each polynomial exactly over the corresponding subinterval Effectively these rules compute the exact integral of a function that interpolates the values of the integrand f 1 on 11 based on the values of f at the endpoints of each subinterval Interpolatory quadrature rules that use k equally spaced sampling points for each polynomial approximation have the property that they obtain the exact result if the integrand is a polynomial of degree k or less if k is odd and degree k 1 or less if k is even For example Simpson s Rule is exact for cubic polynomials since it uses three nodes in each approximation However much higher accuracy can be obtained if we do not require the sampling points to be equally spaced This is the basic idea behind Gaussian quadrature in which the sampling points are chosen so to obtain much accuracv possible Using this approach with k sampling points for each polynomial approximation yields exact results for an integrand that is a polynomial of degree 2k 1 or le swnearly twice the degree that can be obtained using equally spaced sampling points It is for this reason that Gaussian quadrature is often the preferred approach for approximating integrals numerically For more information on numerical methods for approximating integrals well derivatives see http www math uci edujlambersmatl1105a fa390 mm 2W2 4f 3 444 2f 1272 4fw71 WM 6 Example 1 The following table describes a function 3t that indicates the rate of snowfall at time f where t is measured in hours since some initial time MIN 5 02 L41 1 0 12 1 32 2 t 07 12 23 17l11 0 Estimate the total snowfall over the threehour period for which data has been gathered Solution We will estimate the total snowfall which is given by the definite integral 3 m It using five different methods left endpoints right endpoints the Midpoint Rule the Trapezoidal Rule and Simpson s Rule l V 1 Left endpoints the integral can be interpreted not only total snowfall but also the area under the curve 9 91 from t 0 to t 3 This area can be estimated by approximating the region under the curve by rectangles that are determined follows the interval 0 is divided into subintervals of equal with At 3 06 12 Each subinterval has endpoints ti1ti where ti iAt for i 0 6 Then we have 3 3 m tmt Z m dt 8 0 i1 i1 and each integral in the above sum can be approximated follows 5 m at a mam a may 9 ti l In other words the integral of the subinterval ti1 ti is approximated by computing the area of a rectangle with width At and height equal to the value of 91 at the left endpoint of the subinterval Since there are 6 subintervals each of width 12 it follows that the approximation LG to the total snowfall obtained using left endpoints is given by 3 LG ZY C71A i1 3 Zy i 1A1At i1 3 ZrW llElE i1 1 0 1 12 1 1 1 2 07 12 23 17 11 05 lewwlwmlw A l m V l m 10 The approximating rectangles are shown in Figure 1 L511 endpoints n c 39m 1 r0 Figure 1 Approximating rectangles from using left endpoints 2 Right endpoints we proceed exactly with left endpoints but this time we approximate the integral over each subinterval by the area of a rectangle whose height is obtained by evaluating t at the right endpoint ti of each subinterval ti1ti The approximation R6 03 obtained using these right endpoints is Rs 3 2 may Z Z riAtAt gt Z 9i22 Ti 1 12 1 1 2 12 23 17 11 05 02 dedemhd A l V 1 11 The approximating rectangles are shown in Figure 2 Midpoint Rule the area under 9 2 Mt is once again approximated by computing the areas of rectangles of equal width but the height of each rectangle is determined by evaluating 3t at the midpoint of each subinterval Because we only know the values of t at select points we can use at most 3 subintervals of width At 3 03 1 Our subintervals are 01 12 and 23 with midpoints 12 32 and 52 respectively The approximation ME is given by v M3 TWP 1231 H v W WWW i1 3 Era U2 i1 1 121 323 52 121705 34 12 The approximating rectangles are shown in Figure 3 Alternatively the Midpoint Rule can be viewed approximating the region under 91 on each subinterval 174 ff by a trapezoid of A Right endpoints l 39m w r0 Figure 2 Approximating rectangles from using right endpoints width At and heights 1 tMAt2gtquot tM and 9tM At2gtquot tM where tM ti1ti2 is the midpoint of the subinterval Of course we do not know 939quot t M but we can approximate it using the finite difference 9quot t 9quot t flag 1 1 I 13 The approximating trapezoids are shown in Figure 4 Trapezoidal Rule The interval 0 is once again divided into subintervals of equal width At 12 On each subinterval ti1ti the region under 9 9t is approximated by a trapezoid of width At and heights determined by the value of 9t at the left endpoint tFl and at the right endpoint ti The approximation 7 obtained using this approach is 6 A M A Ta ZAt millg39Hm i1 r0 Midpoints 39m Figure 3 Approximating rectangles from Midpoint Rule PMHMHMH 3 5 quotC4 rm i1 rforh f1v f2 f2rfa quot13 quot14 F quot14 quot159 quot15 quotfem We 2w 2m 2w 2w ms rm m7mrm2am2u 2an2mmum m7244 342210m A H 1 CR V 14 6 m m Midpoints l r0 Figure 4 Approximating trapezoids from Midpoint Rule The approximating trapezoids are shown in Figure Simpson s Rule whereas the approach of using left 1 i or right 1 otpplui imam the integrand t by a constant function on each subinterval and the Midpoint Rule and Trapezoidal Rule approximate 9t by a linear function on each subinterval Simpson s Rule approximates 3t by a quadratic function on each pair of adjacent subintervals and integrates each quadratic function exactly using the Fundamental Theorem of Calculus The resulting approximation Sc is 3 S gh ao 41 11 ENQ 4 f3 21 14 4 f5 2916 I f7 1 2 07 412 223 417 211 405 02 1 60748466822202 Trapezoidal Rule l 39m 1 r0 15 The approximating quadratic functions are shown in Figure 6 Note that on the third and fourth subintervals the quadratic approximation is actually a linear function since the points 1 23 El217 and 211 all lie on the same line C Example 2 Approximate the integral 3 x2 11 16 l 0 using the following methods 1 Left endpoints with 3 and 6 subintervals Simpson s Rule l 39m r0 Figure 6 Approximating quadratic functions from Simpson s Rule F0 Right endpoints with 3 and 6 subintervals 9393 The Midpoint Rule with 3 and 6 subintervals r The Trapezoidal Rule with 3 and 6 subintervals Simpson s Rule with 2 and 4 subintervals m Compare to the exact answer 3 3 lt3 x2 11 i o 5 Solution In the case of 3 subintervals we have width A1 3 03 1 and for 6 subintervals we have A1 3 02 12 9 17 0 5 1 Left endpoints We have L3Axa3xx 10212220145 18 11 and LG 2 A1 1 t t t t t NU VVQ VTw 1 0141944254 1 3 2 6875 19 2 Right endpoints We have RgAxp x 1u 22 u49ym em and R6 A1 lt12gt21 lt32gt222lts2gt2321 1 1419442549 191 24 11375 21 3 The Midpoint Rule We have 2113 A1 1le g 1 052 152 25 875 22 and Me 1 52 J j2 Jig2 112 1 i 0252 0752 1252 1752 2252 252 117 87 2 1 89375 23 4 The Trapezoidai Rule We have 2 2 2 2 2 2 2 2 z 73 2 1021I212J3 2 0 21225 202899121 12 Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 16 Notes These notes correspond to Section 104 in the text Exponential Growth and Decay In many applications it is necessary to solve the problem of finding the function 91 that satisfies the simple di erential equation dy i 1 d M where k is a known constant and the initial condition 90 90 2 where yo is a given initial value Together the differential equation and initial condition specify an initial value problem In words the initial value problem states the following o The unknown quantity 91 initially has the value yo 0 At any time t its rate of change with respect to time is proportional to its value at the time t with k being the constant of proportionality We can solve the initial value problem using the natural logarithmic and exponential functions If we divide both sides of equation 1 by y and then integrate both sides from 0 to t we obtain 5 1 dz 5 ii 15 lids 3 r 0 98 18 r 0 Then we make the substitution u 95 which yields du dyds 15 Then this equation becomes W 1 5 7 la lids 4 M u l 0 Evaluating the integrals on both sides yields mm In in it a which can be rewritten ln wit let yo We then exponentiate both sides From the cancellation equations and the laws of exponents it follows that WM hole 7 Now suppose that 90 gt 0 Then we must have 91 gt 0 for any 1 because otherwise by the continuity of 91 we must have 91 2 0 for some 1 which is impossible because a gt 0 for all 1 Similarly if 90 lt 0 then 91 lt 0 for any 1 Therefore 91 and 90 are the same sign so we can drop the absolute values in equation 7 and obtain 91 906 8 If k gt 0 then 91 grows rapidly in absolute value over time and the differential equation is called the law of natural growth In this case the equation can be used a simplified mathe matical model for population growth for example If k lt 0 then 91 rapidly decays to zero over time and the differential equation is known the law of natural decay This law describes for instance the process of radioactive decay Finally if k 0 then there is no growth or dec and 91 remains constant for all time Population Growth The differential equation IF 7 kP 9 11 can be used to model the growth of a population denoted by P1 The proportionality constant k also known the relative growth rate can be determined experimentally by birth rates and death rates since the number of births and deaths per unit of time tend to be proportional to the current population Therefore if the birth rate determined to be some value 13 and the death rate determined to be D then one model for population growth could be IF 7 B D P 10 d lt gt lt gt which would imply P1 Poel m where 1 0 denotes the current time If the population of a region is being measured then other factors such immigration and emigration due to economic forces could also be measured and included in the determination of the relative growth rate Example 1 In 1900 the world population approximately 1650 million In 1910 the population had grown to approximately 1750 million Determine the relative growth rate of the population Solution The population at time 1 where1 is measured in since 1900 is denoted by a function P1 that satisfies the differential equation 1P 71 11 d k H where k is the relative growth rate we are seeking and the initial condition P0 1650 This initial value problem has the solution PU 16506 Substituting 1 10 into PU yields 16506 1750 12 which can be solved for k using the cancellation equation lnex 1 1750 1750 1 1750 10k 3 7 2 1 l 2 7l a z 4 1 1650 0k n 1650 k 10 n 1650 0 00188 J C Radioactive Decay The mass mt of a radioactive substance satisfies 1m 7 r 14 It km where k lt 0 By measuring mt at different times the relative decay rate denoted by k can be determined experimentally since mt mow where mo is the initial mass Then this rate can be used to determine how much of the substance will remain after any given length of time Example 2 The halflife of Radium226 is 1590 Determine the relative decay rate Solution The halflife of a radioactive substance is the amount of time that is nec y for a sample of a substance to lose half of its original mass If mt denotes the mass of a sample of Radium226 t after the initial time t 0 then mt satisfies the differential equation 1m 7 km 15 It gt where k is a negative constant and M denotes the relative decay rate we are seeking The function mt also satisfies the initial condition m0 mg where mo is the initial mass The solution of this initial value problem is mt mow Substitutingt 1590 it follows from the definition of halflife that r m quot1061090quot g 16 ZISQOk 17 Using the cancellation equation lnex J and the laws of logarithms we obtain 1590k ln ln1 hr lr12 18 Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 19 Notes These notes correspond to Section 81 in the text Integration by Parts Recall the Product Rule for differentiation 1 alfU M l fJ39y39J39 y f wl 1 Like other differentiation rules this rule can be reversed to obtain an integration rule Specifically the product fJgJ is an antiderivative of fJg J gJf J that is f f mm gmf39m dx MW c 2 Rearranging algebraically we obtain the equation Mrx dx MW warm 3 where the arbitrary constant C can be dropped because an indefinite integral appears on both sides of the equation If we let u fJ and it 90 then we have du f J 1a and d1 g J 11 It follows that equation 3 can be rewritten u 11 m ltrdu 4 This rule is known the rule of integration by parts To apply this rule properly to an indefinite integral of the fornl lJ39dJ39 439 L41 V it is necessary to identify the functions that will play the role of u and z in the above rule In other words we must be able to write 110 1a in the fornl u 11 where u and it are chosen so that 1 la can be evaluated more easily than the original integral Typically one chooses u uJ to be some portion of 110 and then sets d1 110 uJ 11 Then it and la can easily be deternlined Example 1 Evaluate an1J Solution We use integration by parts with 1 uan dud1 trJ dtrdw I to obtain y y 1 y andJ Jan 11 Jan 11J Jan J C 8 Cl Example 2 Evaluate tan l 9 Solution We use integration by parts with u tan 1 1 la 7quot 11 it 1 d1 1a 10 well a substitution u x2 1 to obtain 71 71 J tan Jtan J 7111 llF 1 Jtan 1 J 7 idu 2 u 1 Jtan 1 J i ln M C xtail lx ganZFHC 11 Cl Example 3 Evaluate a sinxdx 12 Solution We use integration by parts with u a la a 11 it cosJ 11 sinJ d1 13 to obtain We then use integration by parts again with sin111 x cos111 x cos111 14 u 1 111 3 11 it sin1 11 cos111 1 3 to obtain lea sin1 11 x cos111 1 sin1 lea sin1 11 16 We have obtained the same integral that we started with but with the opposite sign We can therefore add this integral to both sides to obtain 2 3 sin111 1 sin1 1 cos1 C 17 which yields 712 1 sin111 2 sin1 cos1 C 18 We now examine how integration by parts can be used to evaluate a definite integral If we integrate both sides of the Product Rule for differentiation from 1 to b we have 5 d 5 31mm d1 fwy w 91f 1gt r11 19 It follows from Part 2 of the Fundamental Theorem of Calculus that 4 f mm fwy w 91f 1gt 20 or rearranging 4 b 4 fxy xdx mm 9wf w r11 21gt Therefore to apply integration parts to a definite integral of the form lib1111 22 we identify the functions 11 and 1 in the case of the indefinite integral Then we can simply compute ubtrb u1tr1 and evaluate the integral of 131u 1 from 1 to b Example 4 Compute WTZ J cosJ dJ 23 l 0 Solution We use integration by parts with du dJ it sin J 11 cosJ dJ 24 to obtain 2 wrZ 7r2 J cos J dJ J sinJ 0 sinJ dJ 0 l 0 392 JsinJ cosJ 3 7r 7r 7r 0 isini cosi cos 2 2 2 7 E 1 25 A common usage of integration by parts is to evaluate integrals of the form me 26gt where the function fJ can easily be anti differentiated repeatedly is the case with various exponential or trigonometric functions For such an integrand one can choose it J which yields IN mJm l dJ By applying integration by parts in this way m times the power of J in the integrand can be reduced all the way to zero hopefully ielding a simpler integrand Example 5 Evaluate ll J26 dJ 27 Solution We use integration by parts with u J2 du 2J1J it 11 a dJ 28 to obtain J26 dJ J26 2 l flx dJ 29 We then use integration by parts again with i u J du dJ d1 a dJ 30 Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 2 Notes These notes correspond to Section 51 in the text Areas and Distances There are many cases in which some quantity is defined to be the product of two other quantities Some examples are 0 If an object is traveling at constant velocity 1 then the distance d that it travels within a time span of length t is given by the relation 1 W o By definition a rectangle of width to has constant height h The area A of the rectangle is given by the formula A wit o The mass m of an object with volume it and density d is given by the formula m 11 Unfortunately in many applications we cannot necessarily assume that certain quantities such velocity or density are constant and therefore we cannot use formulas such d W directly How ever we can use them indirectly in these more dif cult cases by employing the most fundamental concept of calculus the limit The Area Problem Suppose we wish to compute the area A of a shape that is not a rectangle To simplify the discussion we assume that the shape is bounded by the vertical lines J a and J b the J axis y 0 and the curve defined by some continuous function 1 fJ Where fJ 2 0 for a 3 J g b If the function fJ is not constant on the interval 11 then the shape is not a rectangle so we cannot compute its area by using the formula A wit directly However we can use this formula indirectly We begin by attempting to approximate the shape by a rectangle since three sides of the boundary of the shape do form a portion of a rectangle How should we choose this rectangle We consider this question in the following example Example 1 Suppose that we wish to approximate the shape shown in Figure 1 by a rectangle Certainly the base of the rectangle should coincide with the base of the shape since it is a line segment connecting the points 00 and 2 0 It follows that the width of the rectangle is 2 units Choosing the height how r is more dif cult Figure 2 illustrates the consequences of two choices for the height If the height of the rectangle is determined by the height of the shape on the left side then the resulting rectangle has height 1 so the area of the rectangle is 2 As can easily Region bounded by y0 x0 x2 and yx21 l l l Figure 1 Region bounded by the vertical lines J 0 and J 2 the horizontal line 9 0 and the curve 9 J2 1 In this lecture we will learn how to compute the exact area of this region be seen from the figure this value is too small since a significant portion of the shape lies outside the rectangle On the other hand if the height of the rectangle is determined by the height of the shape on the right side then the resulting rectangle has height 5 so the area of the rectangle is 10 Clear v this value is too large because the rectangle contains not only the entire shape but also a significant area outside of the shape In summary all we know so far is that the area of the shape is a number between 2 and 10 The reason wl 39 neither of these estimates are close to the exact area is that the height of the shape varies significantly between J 0 and J 2 whereas the height of a rectangle is constant Howe er if we examine an small subinterval 0 of 0 2 find that the function f J J2 1 tends not to much within the subinterval Therefore a single rectangle could yield an accurate approximation to the area bounded by y f J J J d and y 0 El The discussion in the preceding example suggests that we can obtain a better approximation to the area of a nonrectangular shape by approximating the shape itself with a number of rectangles Height determined by Ie endpoint Height determined by right endpoint i i u i i 57 7 57 7 47 7 47 7 37 7 37 7 gt gt 27 7 27 7 17 7 17 7 a a 1 i i 1 i i 71 o 1 2 3 71 o 1 2 3 X X Figure 2 Attempts to approximate the shape from Figure 1 by a rectangle In the left plot the height of the rectangle is the height of the left side of the shape whereas in the right plot the height of the rectangle is equal to the height of the right side of the shape instead of just one rectangle Then we can use the formula A wit to compute the area of each rectangle and add these areas together We begin bv dividing the interval 0 1 into n subintervals of equal width A1 b a n These subintervals ha endpoints x044 1mm MAJh where xi aHAx39 fori 012 n Then fine n rectangles follows for each i 1 2 n the ith rectangle has a base de ned by the line segment connecting the points 1540 and 130 It follows that each rectangle has a width of Ax since for i 1 2 n 13 1571 aH39A a i 1AJ i 1AJ Ax 1 We can choose the height of each rectangle using a similar approach with a single rectangle setting the height of the rectangle equal to the height of the shape at some point Which point should we choose We consider the first rectangle whose base is the line segment connecting 10 to a Am 0 The purpose of this rectangle is to approximate the area of a portion of the shape This portion is the region bounded by the vertical lines J a and J a Am the horizontal line 9 0 and the curve 9 f 1 To approximate the area of this portion with the area of the given rectangle we can proceed before and set the height of the rectangle equal to the height of either the left side or the right side of this portion of the shape For concreteness we choose the right side in which case the height of the rectangle is fa Ax or f m Therefore the area of this first rectangle is f JAMJA Proceeding in this fashion with the other rectangles we choose the height of the ith rectangle to be for i 12 n Then the area of the ith rectangle is From the areas of these n rectangles we obtain an approximation A for the area of the shape AaAn fJ1AJfJ2AJw fJ39nAJ39 2 Example 2 Figure 3 illustrates this approach to approximating the area of the shape shown in Figure 1 using n 4 and n 8 rectangles In the case where n 4 the width of each rectangle is 2 04 12 and the heights of the rectangles are given by f12 and where f 1 x2 1 Computing the areas of these four rectangles and adding them together we obtain the value 5quot In the case where n 8 the width of each rectangle is 2 08 14 and the heights of the eight rectangles are equal to fi4 for i 1 2 8 Computing the areas of these eight rectangles and adding them together we obtain the value 3 As can be seen from Figure 3 in both cases the approximate area that we have computed is too large because the union of the rectangles contains the entire shape well some additional area However these estimates of the area are much more accurate than our previous estimates of 2 and 10 that were obtained by approximating the shape with a single rectang e C Sigma Notation For convenience we will write summations such the expression for An in equation 2 using sigma notation H An fa1m man Z mom 3 i1 The E1 symbol indicates that the expression to the right in this case fJ39iAJ39 is to be evaluated for certain values of some index variable and the resulting values are to be added The index variable is indicated below the E in this case the index variable is i The starting value of the index variable is also indicated below the E in this case the starting value is 1 The index variable counts from the starting value to the final value which is indicated above the E in this case the final value is n In summary the index variable i assumes the values 1 2 n For each of these values of i the value of the expression f 1AJ is computed and all of these values are added together 4 rectangles 8 rectangles u t u l 5 7 7 7 5 7 i 7 4 i 7 4 7 i 3 7 7 3 7 7 gt gt 2 7 7 2 7 7 1 7 7 1 7 7 G G 1 l l 1 l l 1 0 1 2 3 1 0 1 2 3 x x Figure 3 Approximation the shape bounded by the curve 1 x2 1 and the lines 9 0 J 0 and J 2 using four rectangles left plot and eight rectangles right plot In general sigma notation is useful for concisely der ribing the sum of consecutive elements of a sequence am am1am2 an where m and n are integers Using this notation the sum of all of these elements can be written n 2 ari 2 am 1171 am2 an i 17 The index variable i specified below the El effectively counts from the starting value m to the final value n For each value of 139 between m and 71 including m and n the value of a is included in the sum It should be noted that the letter i is not always used the index variable though this is the most common choice It should also be noted that n need not be greater than m If m n then the summation includes only one term which is am If m gt n then the summation does not include any terms and therefore the value of the sum is zero Computing the Exact Area We now return to the problem of computing the area A of a nonrectangular shape If we approxi mate the shape by n rectangles using the approach described previously we obtain the approximate area An given in equation 3 As can be seen from Figure 3 we can obtain a more accurate ap proximation by choosing a larger value for n Why is this the case The answer lies in the fact that each rectangle is used to approximate the area of a portion of the original shape Since each rectangle has the same width A1 I an it follows that n increases each rectangle is associated with a smaller portion of the shape The smaller the portion is the less that the height of the portion can Therefore the area of the portion can be approximated more accurately by the area of a rectangle This discussion suggests that we can obtain the exact area of the shape by computing the limit of An n becomes infinite assuming that this limit exists This is in fact the case when the function f 1 which defines the top boundary of the shape is continuous on the interval 21 This leads to the following definition De nition 1 Let f 1 be a continuous function on an interval 21 and assume that f 1 2 0 on 21 The area A of the region bounded by the vertical lines J a and J b the horizonal line 9 0 and the curve 1 fU at V n A 7331010 f739jAJ 439 where A1 I an and 1 a iAx for i 12 n We now illustrate how this definition of area can be used to compute the exact area of a non rectangular shape Example 3 We will use Definition 1 to compute the area A of the shape shown in Figure 1 In this case we have a 0 I 2 and f 1 x2 1 We have 7 A l A R 31010 fwd J 7 quot 2 0 2 332 w 1 Th 2 2 nlim 72 1 0 iAa 1 in 2 2 2 l 397 1 7 W 392 41 2 glsozllvrll 7 W 392 81 2 7 812 2 822 2 8m 2 lim 3 7 K 7 4444 3 new 71gt 71 71gt 71 71gt TI 8 2 2 limj1222 mn27H 7 6 00 77f n n To compute this limit we first note that 271 is being added n times which yields n2n 2 Second we use the formula nn 12n 1 1222 n2 6 7 Using the Limit Laws 72131010 a a 7331010 u 072131010 f n 72131010 f n 900 7331010 f n 7331 JUL 8 where a and c are constants we obtain 39 39 1 239 1 A 1m g TE MX 77gt 6 2 Hm nn 12n 1 6 TE MX n3 2 r 2 4 Hm 2n n 1 J ngt 7 4 1 27 lim 2 3 TE MX n n2 4 2 72 3 4 6 9 C The Distance Problem Just the formula A wit for the area of a rectangle cannot be used directly to compute the area of shape whose height is not constant the formula 1 W cannot be used directly to compute the distance that an object travels during a given interval in time if the object s velocity is not constant However we can use this formula indirectly just we used A wit indirectly to solve the area problem Suppose that an object is traveling at a non constant velocity If the object starts moving at time t a and continues until 1 b then what is the total distance traveled For concreteness we let 131 be a function that indicates the object s velocity at time t where a g t g I Then we can approximate the distance by assuming that in very small intervals of time the velocity is nearly constant This is a reasonable assumption if the function 131 that represents the velocity is a continuous function which we will assume in this case We then divide the interval 11 into n subintervals of width At I an with endpoints nihti where ti a iAt for i 012 n It follows that between time 7 71 and time ti for t 1 2 n the distance traveled is approximately MMAt and the total distance d is therefore approximated by the sum of these n distances which we denote by In 1 a In 1311At 1312At 4 4 4 13tnAt 10 Proceeding in the area problem we can de ne the total distance traveled to be the limit of the approximation In n approaches infinity 7 d 7331010 MMAt 11 Example 4 The velocity of a ball thrown straight up with an initial velocity of 100 ft s is described by the function 131 32t 100 where t denotes the number of seconds that have elapsed since the ball has been thrown and 131 is the velocity at time t measured in ft Using the above definition of distance traveled we an compute the distance that the ball travels in the first two seconds after it has been thrown by dividing the interval 0 2 into n subintervals of width A 2 0n and computing the sum 2 n 2 d 2 eff 32tA1007 ts0iAt2i n i12n 12 tch 21 ltgt 1 1 It follows that the total distance d that the ball travels in its first two seconds of flight is d nlglolodn R 2t 2 l 327 100 7 R 1281 200 nlglo o72 7 7 128 4 1 200 128 4 2 200 128 4 n 200 hnl 444f474447 44lt47447 44lt47447 naoo n2 n n2 n n2 n Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 4 Notes These notes correspond to Section in the text The Fundamental Theorem of Calculus In order to evaluate a definite integral we must compute the limit of a Riemann sum which we have seen can be very tedious In this lecture we will discover a much more ef cient method for evaluating definite integrals that will prove to be useful in many Our discussion will be verv similar to the derivation of differentiation rules By the definition of a derivative the derivative of a function fJ at a point J J0 is a number denoted by f J0 and that number represents the instantaneous rate of change of y fJ with respect to J at J Jo In order to compute the derivative using the definition it is necessary to compute a limit which can also be rather tedious To alleviate this difficulty we choose to view the derivative of fJ a function in its own right That is we can define the function f J to be the function whose value at eve y point J0 in the domain of f is equal to the derivative of f at J0 assuming that the derivative exists at J0 Specifically f is defined by W 2 m m h mi 1 hat h for each J in the domain of f By allowing J to be a variable instead of a fixed number we are able to use the definition of the derivative to obtain functions that represent the derivatives of certain types of functions such polynomials or trigonometric functions instead of having to use the definition directly every time we want to compute the derivative of a function at a specific point Example 1 Consider the function fJ J2 We can compute the derivative of fJ at J 2 by using the definition of the derivative to obtain 2 l N gt h 2 4 lim h hat h 4 4h h2 4 lim hyao h lim4h hat 4 2 Alternatively we can use the definition of the derivative in a more abstract manner allowing J to instead of fixing J 2 This yields fWlm Li hat It 2 MM hat It 2 hat It 2 limEJt hat 2J Then we can easily compute f J at any value of C By being even more abstract and using the definition of the derivative with types of functions instead of specific functions we obtain the various differentiation rules with which we are familiar We will use the same approach to simplify the process of computing definite integrals First we will view the definite integral of a function over an interval 11 a function in its own right and try to understand this function s behavior It turns out that this perspective will lead to a remarkable discovery concerning the relationship between integrals and derivatives Part 1 Differentiation Undoes Integration Let f J be a function that is continuous for all J 2 0 How does the definite integral of f change the limits of integration change It is our hope that the answer to this question will provide us with a more ef cient method for computing definite integrals To answer this question we view the definite integral a function of the limits of integration which means that the limits are now variables instead of fixed numbers For simplicity we will only allow the upper limit to and we will keep the lower limit fixed at 0 We define 1 m m at a gt o 4 i 0 Then the value of at any point J0 gt 0 is the definite integral of from 0 to J0 How does gJ change J changes We can answer this question by attempting to compute g J which is the instantaneous rate of change of t gJ with respect to Using the definition of the derivative and the properties of definite integrals introduced in Lecture 3 we obtain 110 11 10 I 2 W 3335 h 1 1 1 l 7 35 NW WW L41 V 1 1 lim 7 t It 39 hat h x To compute this limit we note that if It is suf ciently small then is continuous on the interval ll We may assume that h is suf ciently small because we are computing a limit ll approaches 0 By the Extreme Value Theorem assumes a maximum value and a minimum value on the interval J It Let the maximum and minimum values of f on J 11 be attained at the points MM and mh respectively That is fmlSf ls MUlll x9 Jh 6 Then by the properties of definite integrals 1 mmms imms Mmm lt l V since It is the width of the interval J ll Because we are computing the limit ll approaches 0 we are not concerned with the case of h 0 Therefore we can divide through by h and obtain 1 1 fmmns meg Mw a t 1 Now we let h approach 0 Because J g mh 3 J h and J g M ll 3 J ll it follows from the Squeeze Theorem that li39l M39 9 m l J l J Because f is continuous on ll it follows from the definition of continuity that gymw mm gymmw4w am By the Squeeze Theorem it follows from equations 8 and 10 that 1 1 h l 7 11 gh fmm u gt We conclude that 1 1 Jwrumif imm m um hat h x In words this equation states that the instantaneous rate of change of the area of a region with respect to its width is equal to the height of the region at the point where the width is changing This is to see in the simple case of a rectangle of width 11 and height ll Since the area A of the rectangle is given by the formula A wit we can easily determine that the rate of change of the area with respect to the width is given by 1A dwh 13 It 110 h since the height of a rectangle is constant Now however we can compute the rate of change of the area of a region with respect to its width even if the height is not constant More generally suppose that we define the function F 1 by ox M mm 14gt r a where ft is continuous on 0 b Then Part 1 of the Fundamental Theorem of Calculus states that for any 7 in al F is continuous and differentiable at 1 and F The function F 1 is called an autideriuatiue of f 1 Later in this lecture we will see how antiderivatives can be used to easily evaluate definite integrals Example 2 Differentiate the function Fm Shle 1 It 15 Solution Recall Part 1 of the Fundamental Theorem of Calculus which states that if f 1 is continuous and the function F 1 is defined by the integral 1 M mm 16gt l a then F 1 f 1 Note that the value of F 1 at any J is the area under the curve 9 ft between t a and t Applying this result to the given function F 1 we have F J sint2 1 1 sinJ2 1 17 Cl Example 3 Differentiate the function FJ sint2 1dt 18 Solution Let gJ be the function defined by 1 11 sin12 1m 19 1 3 From the previous example we know that g sinJ2 1 The function F in this example can be written 90 since F 1 and 91 are identical except that the upper limit of integration for is J and the upper limit for F is 12 Therefore F can be obtained from simply by replacing 7 with 12 in the definition of We can differentiate F 1 912 using the Chain Rule which yields 1 1 1 1 F39J awwz g J22J SillJ22 121 21 sinJ4 1 20 Cl Example 4 Differentiate the function 12 FJ sint2 1dt 21 1 11 Solution Let gJ be defined by 1 sintz 1 It 22 1 0 Then in the previous example 9 1 sinJ2 1 Using the properties of definite integrals we can write F 1 12 FJ sintz1dt 1 11 0 A 1 A 51111er 1dt sint2 1W 1 11 1 0 911 12 sint21dt manor 1 0 1 0 y1J 903 23 Therefore we can use the Chain Rule to obtain 1 1 d F39 7 2 7 1 lt1 mm gt1 UM m1 y x2gt2x39gt y 1xgt 1xzgt mm 1x21 sum1 1mm sin 17 1 1 278111J4 1 Example 5 Differentiate the function Solution We define the function 60 by 1 so aw am i 0 By Part 1 of the Fundamental Theorem of Calculus Gquot Using the same approach in the previous example we rewrite F 1 using the properties of definite integrals and obtain 0 111 PM AWfUWfo mm W m immf fmm i0 0 Ghltxgtgtaltgxgtgt 27 Using the Chain Rule we obtain F m gaunt gamma 39ltgltxgtgtg39ltxgt a wmm m fltgltxgtgtg39ltxgt fltnltxgtW 2 C Part 2 Integration Undoes Differentiation Previously we learned that the derivative of the integral of a function f 1 with respect to the integrals upper limit equal to Specifically if 1 m mm 29gt l a then by Part 1 of the Fundamental Theorem of Calculus g This result however is only one part of this theorem The second part which follows from the first part described in equation 29 tells us how we can evaluate the definite integral 3mm mm 6 Suppose that F 0 is any antiderivative of f that is F 0 f0 Then the antiderivative 90 defined in equation 29 satisfies 90 F 0 C for some constant C since 9 0 F 0 E 0 implies F must be a constant By the properties of definite integrals and the definition of 90 we have 90 0 It follows that UUMr an 10 10 2 F0 C Fa C Fl F02 In summary we can compute a definite integral of simply by finding an antiderivative of and then evaluating the antiderivative at the limits of integration Statement of the Fundamental Theorem 0f Calculus We are now ready to summarize these results and formally state the most important theoretical result in all of differential and integral calculus In the statement of the following theorem note that we assume that functions are differentiable on a closed interval A function f 0 is differentiable on a closed interval 00 if it is differentiable on al and f is differentiable from right at a and differentiable from the left at I that is the onesided limits f 1 f a f 1 f b 32 J 1 lim J l and lim xgta 1 both exist This is similar to the definition of continuity on a closed interval Theorem 1 The Fundamental Theorem of Calculus Let f 0 be continuous on 0 b 1 If the function 90 de ned by the de nite integral 1 unfma an i a then 90 continuous on 0 b and di erentiable on 0 b and 9 0 f0 a 3 J g l 34 2 If F0 any antideiiuatiue of f0 on ab that if F 0 f0 a 3 J g l 35 then libf0 dJ F0 F0 36 Example 6 Let J2 Then if we define mmJMm w an i 0 l 0 then by Part 1 of the Fundamental Theorem of Calculus g 392 Furthermore if F J3 then F is an antiderivative of since F J2 It follows from Part 2 of the Fundamental Theorem of Calculus that 4 J2 1 H4 F0 7 38 l 0 This implies that the area under the graph of J2 from J 0 to J 4 is equal to 643 El Differentiation and Integration as Inverse Processes In words the Fundamental Theorem of Calculus states that the derivative of an integral of a function is the function itself and that the integral of a derivative of a function is the function up to an additive constant In other words differentiation and integration are inverse processes just division and multiplication are inve se operations This parallel is even closer than it would appear on the surface Consider the following 0 When speed is constant over time distance is the product of speed and time In general distance is the integral of speed over time o If speed is constant over time then it is equal to the ratio of distance to time In general the speed at any given time is the derivative or rate of change of distance traveled with respect to time Computing Area Using the Fundamental Theorem of Calculus Example 7 Let It and I be positive constants Use the Fundamental Theorem of Calculus to compute the area of the region bounded by the curve 1 fJ where f J hJb the horizontal line 9 0 and the vertical lines J 0 and J 1 Solution The region is shown in Figure 1 It is a triangle with height It and a base of length I By the definition of the definite integral the area A of this region is given by b b hJ 2120 f739d7 0 Trix 59 We use Part 2 of the Fundamental Theorem of Calculus which states that for any function f that is continuous on 11 fumWFw Fm mm 8 yhxlb X Figure 1 Region bounded by the curve 1 f 1 ltxb the horizontal line 9 0 and the vertical lines 7 0 and J b where F 1 is any antideriyatiye of f that is F 1 f 1 We therefore need to find an antideriyatiye of f 1 ltxb To accomplish this we use two anti differentiation rules first if F is an antiderivative of then for any constant 0 CF is an antideriyatiye of cf Second the antideriyatiye of 1 for any integer n y 1 is 1 n 1 It follows that if f 1 ltxb then tJ11 LI392 7 41 F0 3 1 1 2b is an antideriyatiye of f 1 This can be veri ed by differentiating F 1 which yields 1 LI392 It 1 i It 11 3977772727 42 F J 11 2b Ebrlwp 21 I y Now that we have found an antideriyatiye of f 1 we apply the Fundamental Theorem of Calculus and obtain b sz 102 1th 1 f fJdJ Fb F0 2b 2b 7b bh 15 which is the familiar formula for the area of a triangle of height It and base of length I C Example Compute the area of the region bounded by the graph of 92 91 ll J 91 44 y f 1 the line 9 0 and the lines J 0 and J ll where yl yg and Il are all constants Solution This region is a trapezoid of width It and heights 91 and yz The function is a linear function defining a line that passes through the points 091 and ll 92 To compute the area of this region we again use Part 2 of the Fundamental Theorem of Calculus Using the rule that an antiderivative of J is W n 1 provided n 7 1 we can determine that the function FJ defined by 2 12 t 1 J FJ 7 le 443 ll 2 is an antiderivative of This can be veri ed by differentiating F in the previous example From Part 2 of the Fundamental Theorem of Calculus we find that the area A of the given region is m A fJdJ in 92 91 i 0 h Fm F0 92 91112 92 91 ll ll JyldJ i91h i910 2 92 91 ll 2 02 2 91 ll 92 91 ylh hi it lu 2J2 2J1 J1 it it 2 J2 2 J1 hm 46 2 which is the well known formula for the area of a trapezoid of width It and heights 91 and 92 C 10 Example 8 Compute the area of the region bounded by the curve 9 sinx where 0 3 J 3 7r and the line 9 0 Solution The region is shown in Figure 2 To compute the area A of this region we need to 15 1 7 ysin x 7 gt 05 G y0 05 l l l 1 0 1 2 3 4 x Figure 2 Region bounded by y sinJ and y 0 evaluate the definite integral 7 A sinJ 11 47 i 0 To accomplish this we first obtain an antiderivative of sinJ Since 739 39 g sinJ 48 d1 new lt gt it follows that F 1 cosJ is an antiderivative of sin 1 can be veri ed by noting that F J cosx icos1 sinx sinx 49 11 1a Therefore by Part 2 of the Fundamental Theorem of Calculus the area A of the region is given by 7 A sinJ1J 0 Frr Fm cos7r cosO 1 1 1 1 2 50 Cl Net Area VS Total Area Example 9 Evaluate the definite integral o27r sinJdJ 51 l 0 What is the geometric interpretation of this integral Solution Using the Fundamental Theorem of Calculus and the fact that an antiderivative of fJ sinJ is FJ cos J we obtain lizxsinde F27r F0 r0 cos27r cos0 1 1gt 11 0 52 The graph of sinJ on the interval 0277 is shown in Figure 3 What is the geometric significance of the integral being equal to 0 Note that in the previous example we determined that the integral of the same function from J 0 to J 7r is equal to 2 and that is the area of the region bounded by the curve 9 sinJ and the lines 9 0 J 0 and J 7r From Figure 3 we can visually determine that the area of the region bounded by the curve 9 sinJ and the lines 9 0 J 7r and J 2r is also equal to 2 Given this information why is the integral from J 0 to J 2r equal to 0 We can answer this question by computing o27r sinJ1J F27r F7r 12 ysinx on 1127 l Figure 3 Graph of y sinJ 0 lt J S 27 cos27r cos7r 1 1 1 1 2 53 This exanlple illustrates that b fJ dJ is not always equal to the area of the region bounded by the curve 9 fJ and the lines 9 0 J a and J 1 Rather it is the net area above the that is area above the J axis is counted positively while area below the J axis is counted negativ Therefore the area between J 0 and J 7r which is positive cancels with the area between J 7r and J 27 which is negative thus explaining why the integral is equal to zero El In general to compute the gross or total area rather than the net area of the region bounded by y f J y 0 J a and J b it is necessary to evaluate the definite integral Wm 13 dJ 54 Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 12 Notes These notes correspond to Section 71 in the text Inverse Functions There are an ever increasing number of applications in which it is necessary to solve problems that are known inverse problems In such problems the traditional roles of input and output are reversed Instead of asking what is the output from a given input one may instead know the output already and the input needs to be determined The original problem is referred to a forward problem while the new formulation is called an inverse problem We are already aware of some inverse problems If one considers the problem of multiplying two numbers the forward problem then dividing is the inverse problem If differentiating a function is the forward problem then anti differentiating is the inverse problem In many scientific and engineering applications the forward problem consists of determining the response to some stimulus The corresponding inverse problem is given a response that has been measured what is the stimulus that produced the response In many cases the solution to some forward problem may be an equation of the form 1 where J is the input to the function and y is the ouptut The corresponding inverse problem consists of finding an inverse function of f which we will denote by f l that can be used to determine the input J corresponding to ai given output 9 When does such an inverse function e t Recall that a function fJ is defined to be a set of ordered pairs of the form J y The element J is in the domain of f which is the set of all possible inputs of f and the element 9 is in the range of f which is the set of all possible outputs of f For f to be a function it must have the property that for every J in the domain there is exactly one element 9 in the range such that i f J For an inverse function f 1 to exist the opposite must be true for each element 9 in the range of there must be exactly one element J in the domain off such that J fquoty Now suppose that a function f J has the property that for two different elements J1 and J2 in its domain f J1 f72 Then f cannot have an inverse function since the value 9 fJ1 in the domain of f 1 corresponds to two values in the range of f l J1 and J2 and therefore f l cannot be a function It follows that in order for to have an inverse function it must have the following property De nition 1 Onetoone A function fJ one to one if for any two distinct elements J1 and J2 in the domain if f fJ1 and fJ2 are also distinct Example 1 The following function is not oneto one since 2 2 Cl Cl Example 3 Let t describe the height of a kicked football at time t This function is not one to one since the football must eventually come back down after gaining altitude immediately after being kicked C There is a simple test that can be used to determine visually if f is oneto one known the horizontal line test If it is possible to draw a horizontal line that intersects the graph of f more than once then cannot be oneto one since the horizontal line corresponds to an element in the range of f that is associated with more than one element in the domain Now that we have determined the conditions under which an inverse function exists we can now provide a precise definition of an inverse function De nition 2 Inverse Function Let f be a onetoone function with domain D and range R The inverse function of f denoted by f l the function with domain R and range D de ned by the condition J f 1y if and only if y fJ39 1 where J belongs to D and g belongs to R Note that the domain of f 1 is the range of f and vice versa This is consistent with our previous statement that when considering inverse problems the traditional roles of input and output are reversed If f has an inverse function how can it be determined This is not always to do but the general procedure is to solve the equation g fJ for J thus obtaining the relation J f 1 g The graph of 1 however is to obtain from the graph of If the graph of is reflected about the line g J then in the new graph the roles of J and g are reversed and the reflection is the graph of f l If the inverse function is dif cult to determine algebraically this graph may be helpful to at least approximate or visualize 1 Furthermore the relationship between the graphs of and f 1 suggests the following result Theorem 1 Continuity of Inverse Functions If f onetoone and continuous at a and b ay then f l continuous at 1 Example 4 Let fJ 1 2J 37 on the domain 13 lt J lt 00 Show that fJ is oneto one on this domain and find the inverse function 1 Solution We have f J 2 6J It follows that for J gt 13 f J gt 0 and therefore fJ is increasing for all J in the given domain This means that it is not possible for fJ to assume a value more than once on this domain though f J is not one to one on the domain 00 lt J lt 00 since it is a parabola To find its inverse we solve the equation y12J3J2 2 for We use the quadratic formula to solve the equation 3J22J1 y0 3 and obtain 2 j 22 431 y 4 1 mm ltgt which simplifies to 1 f 2 3 1 L593 5 Since the domain of is the interval 1 so it follows that the range of the inverse function defined by J f ly must also be this interval Therefore we must choose the positive square root and we have 7 1 M31 2 fww r 4i m 5 5 which if desired can also be written 7 1 2 f 10 3 I Even if a formula for the inverse function is not available it is still possible to determine its derivative at a given point Intuitively if I f a and f o m then the slope of the tangent line of f at J a has slope m and therefore the slope of the tangent line of f l at y I should have slope 1m in other words f 1 b 1m This should be true because reflecting the tangent line about the line 9 J has the effect of taking the reciprocal of the slope the roles of rise and run are reversed where slope is conceptually rise over run To determine if this is the case in general we can rely on the cancellation equations which state that 71 1 ff 9 y and f NJl J 8 for every 9 in the range of and every J in the domain of Taking the first cancellation equation and differentiating both sides with respect to 9 via the Chain Rule we obtain f39f ly39f 139y 1 9 which yields the following differentiation rule Theorem 2 Di erentiobility of Inverse Functions Let f be a onetoone function that di er entioble at and let i fJ If f J 0 then f 1 di erentioble at i and f f l 9 In words this theorem states the derivative of the inverse function is in a sense the reciprocal of the derivative However it should be noted that f must be evaluated at f 1 9 not J It is important to note that the derivative of f 1 is only defined at y if f fquoty y 0 in other words f must not have a horizontal tangent at f 1 y This only makes sense because this horizontal tangent of f would by reflection correspond to a vertical tangent of f 1 at 9 meaning that f 1 should not be differentiable at y anyway f lWJ 10 Example 5 Let J2 1 on the domain 0 Compute f l 2 without computing f 1 or using the differentiation rule for inverse functions Solution Setting J fquot2 and applying the cancellation equations yields fJ J2 1 2 which has the solution J i1 Since the domain of for this problem is 0 so we must have fquot2 1 At J 1 the slope of the tangent line to the graph off is equal to 2 since f J 2J which yields f 1 2 The graph of f 1 can be obtained by switching the J and y coordinates of points on the graph of so this line that is tangent to the graph of at the point 1 2 corresponds to a line that is tangent to the graph of 1 at the point 21 The tangent line for 1 therefore must have slope 12 since slope is defined to be the ratio of change in y to change in J and the roles of J and y are reversed in graphing 1 or any of its tangent lines The graphs of and 1 along with the graphs of the line tangent to f at 12 and the line tangent to f 1 at 21 are shown in Figure 1 E Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 13 Notes These notes correspond to Section 72 in the text The Natural Logarithmic Function De nition of the Natural Logarithmic Function Recall that the power rule for integration 17 1 d1 C 1 n 1 does not apply if n 1 since then there is a division by zero However since 1 t is continuous for all t gt 0 then its definite integral exists over any interval 21 where I gt a gt 0 We can therefore de ne a function of J to be the integral of 1 t over an interval whose right endpoint depends on De nition 1 Natural Logarithmic Function The natural logarithmic function an de ned by ix 1 an dt Jgt0 2 f 1 Note that this function can easily be differentiated using Part 1 of the Fundamental Theorem of Calculus d d x 1 1 l 739 7 If 7 3 11 x 11 t J From the chain rule we obtain the following more general differentiation rule Theorem 1 If 9a a positive di erentiable function then y U39 10 grim We now illustrate the usage of this rule with some examples Example 1 Compute the derivative of lnsin sinln Solution Using the Product Rule and Chain Rule we obtain sinanlnsin lnsinJsinln 1 cosJ lnsinJ cosan7 sinJ J sinln l 39 u l sinan cotJ 5 Cl Example 2 Differentiate f 1 lnJquot ln 1 Solution We have 1 1 1 7 7 1I773571 7 6 f Jquot anJ my Jquot an w 1 Cl Example 3 Differentiate f 1 ln1 1 Solution FI OXH the Chain Rule we have 1 21 I 2 39 7 fm 1H 1H2 H C Properties of the Natural Logarithmic Function The natural logarithmic function has some interesting properties all of which can be proven by applying the chain rule for differentiation lnJy an lny 8 ln an lny y lnJT 9quot an 10 In addition an has the properties lirn an so lirn 30 11 xgtoo zoo Furtherrnore because its derivative is always positive for J gt 0 it can be seen that an is oneto one and therefore has an inverse function Example 4 Differentiate 2 J 1 12 f0 n7271 Solution Using the Chain Rule and Quotient Rule we obtain it J2J 1 J2J1J2 J2J2J1 J2 J2J12 J2J 1 J2J12J J22J1 J2 J2J12 J2 J 12J 2J 1 2 2J1 J2J1 15 Another way of obtaining the derivative is to use the laws of logarithms in advance Writing fJ an2 lnJ2 J 1 2an lnJ2 J 1 14 we obtain 2 1 2 2 1 2 J J1 J I v 7 1 f0 J J2J1 J J2J1 J C Integration Rules In a sense the definition of ln J can be extended to include negative J by working with ln which is equal to ln J for J lt 0 Note that i J gt 0 Mn 1 16gt fig 1 J lt 0 This yields the differentiation rule 1 1 alll 1 and the following integration rule Theorem 2 The inde nite inteng of the function 1u i 1 a In lnu C 18 where C an arbitrary constant This rule can be used to address a conspicuous void in the list of integration rule we have developed While we have simple rules for antidifferentiating sin1 and cos 1 we do not have such rules for the other four basic trigonometric functions We can now derive such rules the following example illustrates Example 5 Compute tan1d1 19 It follows Solution Writing tan1 sin 1 cos1 and letting it cos 1 we have du sin1 11 that sin1 tan 1 d1 11 cos 1 n u lnu C C 10 lncos1 ln cos1 C C 20 ln C08 7 ln sec1 As will be seen in later lectures similar integration rules can be obtained for cot 1 sec 1 and csc1 Cl Logarithmic Differentiation The natural logarithmic function can be used to more easily differentiate functions that are com plicated products or quotients since logarithms of products and quotients are equal to sums and differences For example if g1t1 then ln lng1 lnt1 21 Differentiating both sides we obtain f U39 M 22 which can then be solved for f 1 Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 18 Notes These notes correspond to Section 77 in the text Indeterminate Forms and I Hospital s Rule Suppose that fJ and gJ are two differentiable functions such that fa 10 0 and both derivatives are continuous at J L Then it is not clear whether L lim 1 xgta exists since it depends on the rates at which fJ and gJ approach zero J approaches a However because both functions are differentiable it follows from the definition of the derivative a 2 that L lim lim limina f39far Ha I 3 3 glm 5 In summary we can determine the limit L of g J gt a by differentiating both functions and computing the limit of f g If f a g a 0 and both derivatives are themselves continuously differentiable at a we can repeat this process until the limit can be determined This technique of computing the limit of the quotient of the derivatives is known I Hospz39tal s Rule which we now state Theorem 1 I Hospital s Rule Let f and g be functions that satisfy the following conditions 0 f andg are di erentiable at a and g J 0 for J near a except possibly at J a itself and 0 either giro 0 and 11331119 0 4 or fJ Eo0 and gJ 00 5 Then I lim lim 7 Ha 10 Ha 9 0 It should be noted that this rule applies whether the limit L itself is either finite or infinite If f and g approach 0 J approaches a we that limxna is an indeterminate form of type 00 On the other hand if both functions approach either 30 or o0 then we that limxwl is an indeterminate form of type 0000 Example 1 Compute sin J lim 7 7 xgt0 J Solution This limit is an indeterminate form of type 00 Applying l Hospitalquots Rule yields lim mm lim 1 8 xgt0 J xgt0 1 Cl Example 2 Compute J l 7 1 9 Solution This limit is an indeterminate form of type 3000 Applying l Hospital s Rule yields 1 1 1 i 10 1322x21 0 lt gt C Example 3 Compute lim 11 100 1 Solution This limit is an indeterminate form of type 3000 Applying l Hospital s Rule yields 1 lim lim 5 0 12 xgtoo 31 xgtoo 31 Example 4 Compute lim 7 r xgtoo xi1 15 Solution This limit is an indeterminate form of type 00 00 Applying l Hospital s Rule yields 1 x2 1W2 J 14 J lim lim u lim xgtoo 1 2 xgtoo 12J2 1 122x xgtoo This limit is also an indeterminate form of type 00 00 Applying l Hospital s Rule again yields 2 12 2 712 1 1 lim Haw 1 21 lim J J xgtoo 1 xgtoo x2112 lim 15 100 Unfortunately we are back where we started so we cannot evaluate the limit by applying l Hospital s Rule directly We can instead use the fact that 2 2 i i 1 39 13 2112 llggo2112 6 to conclude that J 1392 12 V 1 x2112 JZ1 I The limit on the right side is an indeterminate form of type 00 00 Applying l Hospital s Rule yields 2 lim lim 7 xaooxzI l xgtoo 274 1 This limit is also an indeterminate form of type 00 00 Applying l Hospital s Rule again yields 18 1 2 r 2 1 19 xggoxzI l xgloloi It follows that J 1392 12 v 112 1 2 1131010 2 112 x2 1 0 Other indeterminate forms can be evaluated using l Hospital s Rule For example suppose that J gt a f 1 gt 0 and 90 gt 00 Then the limit L gigllszMW 21 is an indetemnmate form of type 0 00 By rewriting the product a quotient flg or glf we obtain an indeterminate form of type 00 or 0000 in which case l Hospital s Rule can be applied Example 5 Compute limsinJan 22 xgt0 Solution This limit is an indeterminate form of type 0 so We can write it a quotient using sinJ 1 cst and obtain an 23 xao cst This limit is an indeterminate form of type 3000 Applying l Hospital s Rule yields an 1 J sin2 J lim 7 24 7 i 1 xgt0 cst xgt0 cst cotJ xgt0 JcosJ39 using the fact that cotJ cos J sin This limit is an indeterminate form of type 00 Applying l Hospital s Rule again in conjunction with the Product Rule yields sin2 J 2 sin J cos J lim ii 23 xgt0 J cosJ xao cosJ J sinJ Cl Similarly if and both become infinite J gt a then the limit L l 26 123ny 91 gt is an indetemnmate form of type so so This difference can be converted into a quotient using techniques such computing a common denominator rationalizing or removing a common factor The result is a new indeterminate form for which l Hospital s Rule is useful Example 6 Compute 1 1 lim 7 7 2 1a an J 1 Solution This limit is an indeterminate form of type so 00 Finding a common denominator we can rewrite this limit 1 I J nJ 28 l anJ 1 This limit is an indeterminate form of type 00 Applying l Hospitalquots Rule yields 1 l 1 1 lim lim 1a anJ 1 1a an Multiplying the numerator and denominator by J yields 1 1 N 30 1 J 1J an 1 4 This limit is an indeterminate form of type 00 Applying l Hospital s Rule again yields 1 J 1 I 1 1 im 2 13311 1Flnj 1amp111J 2 the llllllt 91 l 31 3 can be an indeterminate form of type 00 000 or 100 depending on the limits of and J amp a In all three cases we can compute L by computing the limit of lnf 19 ln which is an indeterminate form of type 0 4 00 Example 7 Compute lim1 1 1amp0 Solution This limit is an indeterminate form of type 100 We write lim1 1 e 1amp0 where 1 L lim ln1 111 lim amp ln1 1amp0 1amp0 J This limit is an indeterminate form of type 00 Applying l Hospitalquots Rule yields l 1 1 1 L 1gmm 1gmw 1 1amp0 J 1amp0 It follows that V lim1 1 a 61 e 1amp0 Cl Example 8 Compute lim 1 131 1 30 Solution This limit is an indeterminate form of type 000 We write lim 1 1 e 1 30 where 1 L lim ln1 111 lim amp ln1 1amp00 1amp00 J 33 34 38 39 40 Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 3 Notes These notes correspond to Section 52 in the text The De nite Integral In the previous lecture 39e approximated the area under the graph of a continuous nonnegative function y fJ between the vertical lines J a and J I by computing the sum of the areas of n rectangles determined by the division of the interval 11 into n subintervals of equal width AJ I an Intuitively we concluded that n gt so the approximate area N Am Ziaam 1 i1 where 75 1 iAJ for i 1 2 n converges to the exact area of the given region More generally suppose that for each n 1 2 we define the quantity RR by choosing points a J0 lt J1 lt lt Jn b and computing the sum n RR ZfU39DAJ i Axe 1 J iel 17713 14 s 1 2 i1 The sum in equation 2 is known a Riemann sum Note that the interval 11 need not be divided into subintervals of equal width and that fJ can be evaluated at arbitrary points belonging to each subinterval The Riemann sum Rn approximates the area under 9 fJ between J a and J b by the sum of the area of n rectangles where the width of the ith rectangle is AJ and the height is for i 1 2 n This is a generalization of our previous approximations of the area in which we required that each rectangle had the same width AJ 2 b an and for i 1 2 n the point belonging to the ith subinterval Ji1Ji always chosen to be the left endpoint J54 or the right endpoint Ji If fJ 3 0 on 11 then n gt 00 the Riemann sum RR converges to the area under the curve 9 fJ provided that the widths of all subintervals Ji1Ji for i 12n also approach zero If f assumes negative values on 0 b then under the same conditions on the widths of the subintervals Rn converges to the net area between the graph of f and the J axis where area below the J axis is counted negatively This is due to the fact that the height of each rectangle is determined by f 139 so when fJ is negative rectangles will have negative height and therefore negative area This will be discussed in greater detail in the next lecture A Our definition of the area under the graph of f 1 between J a and J b while precise is also unfortunately rather cumbersome This can be alleviated by introducing notation that concisely describes the whole process of computing the area in the manner in which we have described To that end we state the following definition De nition 1 Let the function f 1 be continuous on the interval 0 b Furthermore let RR 530 be a sequence of Riemann sums as de ned in equation 2 that has the property that lim max Ari 0 TE MX 199 We de ne the de nite integral of from a to b to be b M dx 7331010 Rt 4 The function f 1 called the integrand and the values a and b are called the lower and upper limits of integration respectively The process of computing the value of a de nite integral called integration The concept of the definite integral embodies the entire process of approximating the area of a complicated region by reducing the problem to several simpler problems of computing areas of simple regions and then obtaining the exact area through a limiting process By representing the result of such a complex process using such simple notation we can much more easily obtain solutions to a wide variety of problems This is the primary advantage offered by calculus it provides us with a simple means of working mathematically with intuitive concepts such distance and area Evaluating Integrals Now that we have defined the definite integral for the purpose of computing quantities such areas and distances we now discuss the problem of actually evaluating a definite integral of the form 4 M a for a given function f 1 and limits of integration 2 and b Certainly one approach is to use the definition directly This entails approximating the integral using a Riemann sum of the areas of n rectangles and then computing the limit n gt so In the case where is a polynomial it is helpful to use various well known formulas for evaluating summations such it rim 1 i1 n 2 quotn 12n 1 2 l 6 7 Similar formulas for higher powers of i can be derived by solving quottems of linear equations The following summation rules are also useful for reducing the complexity of Riemann sums m 8 R c 2 a 9 i1 R R R 2 117 1 2 117 2 1 10 i1 i1 i1 R R R 20i 11i gas 207 11 i1 i1 i1 We now show how these rules can be helpful for the task of evaluating a definite integral using the de nition Example 1 Compute the area of the region bounded by the curve 1 f 1 where f1 12312 12 well the horizontal line 9 0 the vertical line 1 0 and the vertical line 1 3 Solution The area A of this region which is shown in Figure 1 is given by the value of the de nite integral 3 3 142 f1d112312d1 13 r o l 0 To compute this integral we approximate the region by n rectangles of equal width with height determined by f1 Specifically we divide the interval 0 into n subintervals of equal width A1 3 0n 371 These subintervals have endpoints 1011 1112 Wilma where 1 iA1 for i 0n These points are used to determine the rectangles that approximate the original region The ith rectangle is determined by the ith subinterval 13415 follows the width is the length of the subinterval which is A1 371 and the height is given by 31 2 Figure 2 illustrates these rectangles for the case of n 4 To approximate the area of the region we can compute the areas of the n rectangles and add them The resulting sum is an example of a Riemann sum As H the number of rectangles Area under yx23x2 l l Figure 1 Region bounded by the curve 9 x2 2 and the lines 9 0 J 0 and J 3 increases we can expect that the approximate area will approach the exact area This is in fact the car which is why the definite integral is de ned to be the limit of a Riemann sum of the areas of n rectangles n becomes infinite In other words A b N dx Wiggle mam 14gt 1 i1 where the sigma notation 27 is used to denote the addition of n expressions that are indexed by the variable i that coun from 1 to n We now compute this limit of a Riemann sum to obtain the area of our region Using the laws of sums we have 3 A J23J21J l0 Approximating rectangles n4 Figure 2 Approximating rectangles for the case of n 4 n 31 2AJ 7 n ZKle 31Ax 2Ax i1 n 2 c 31 1 1 i i 2 f n f v 2 f iim i 3 2 00 7 i1 7139 TI 3 quot 3i 2 quot 3i quot r 7 i i 2 20 i1 i1 E n n 92 n 9 7 Zn 2312 3 lim 7 7100 71 lim 7 7100 TI 7 i i 1 1239 1 2 Km i in n 2mm 1 2n ngt00 n n2 6 n 2 2 Km 2 Mn 12n 1 2nn 1 6 THOO 6713 2n2 23923 2 a 2quot r 2 Km n n n n fn 6 THOO 6713 2n2 27 3 1 27 1 lim 27f 7 17 6 Race 6 n n2 2 n 2 lim 2 i lim 1l 6 6 vHoo n n2 2 new n 27 9 7 6 2 285 15 Certainly this process of evaluating a definite integral using the definition is quite tedious In the next lecture we will see how a diligent study of the behavior of the definite integral can lead to much more ef cient methods of evaluation at least for certain types of integrands such the polynomial featured in the preceding example The Midpoint Rule Suppose that we are computing a Riemann sum by dividing 11 into n subintervals 15415 for i 1 2 n where m a and 1 I By the definition of a Riemann sum in equation 2 we can approximate the underlying definite integral by evaluating the integrand at any point in each subinterval 1371 to obtain the height of the corresponding rectangle However there is a particular advantage to choosing to be the midpoint of the subinterval mi 1371 1 2 In the case where f is a linear function on 15415 the area of the corresponding rectangle is equal to the exact area under f 1 since the region bounded by the graph of f the and the vertical lines J 1371 and J 1 is a trapezoid with average height rm This technique of choosing to be the midpoint mi is called the Midpoint Rule For more general functions the Midpoint Rule tends to yield a more accurate approximation than using either the left or right endpoint of each subinterval 611d Properties of the De nite Integral The properties 811 of summations can be used to prove analogous properties for definite inte grals For example by 8 lib odd b a 16 Similarly the rules 9 11 can be used o establish the following additional properties of definite integrals In stating these properties we assume that f 1 and gJ are continuous on the interval 11 and that c is a constant 5 5 cfJdJ 0 fJdJ 17 b a l b fltxgtgltxgtdx fwdx mm us 1 1b 1b fltxgt gltxgtdx fem f max 19gt These properties show that the definite integral is a linear function of the integrand just the differentiation operator ddJ is a linear operator in view of the Sum Rule Difference Rule and Constant Multiple Rule for differentiation There are some useful inequalities pertaining to definite integrals First of all if g then b b fJ 11 g gJ 11 20 l a l a It follows that if m g f 1 g M then 43 mb a g fJ 11 g Mb a 21 t a This property will be v ry useful in the next lecture when we prove a very important theoretical result concerning definite integrals Intuitively the area under the graph of from a to I combined with the area under the graph of f from I to 0 should equal the area under f from a to c More precisely jfJdJ39 fxdxAa x m 22 A related property is that interchanging the limits of integration reverses the sign of the integral ie b a fJ 11 fJdJ 23 a l b This is easy to see by examining a Riemann sum since it includes differences of x values that are negated by interchanging the limits We illustrate this in the following example Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 15 Notes These notes correspond to Section 74 in the text General Logarithmic and Exponential Functions General Exponential Functions The natural logarithniic and exponential functions can be used to define exponentiation using any base a gt 0 Since a can be written 63quot using the cancellation equations it follows that 1 11111 x 1111 a c c 1 since 631W a for any J and y We thus have the following definition De nition 1 General Exponential Function Let a be a positive real number The exponential function with base or denoted by a de ned by ax 6111111 2 where J any real number Example 1 Write xi using the natural exponential and logarithniic functions Solution Since the general exponential function a is defined by a 6 1quot it follows that we can set a xi and J xg and obtain e m By the laws of logarithnis we can eliminate one square root gummy 4 This definition can be used to easily prove the well known laws of exponentiation 11er 1107 5 a 7 a my 6 any a 7 01 all 8 It is not hard to see that these laws follow almost directly from the corresponding laws for the natural exponential function The above definition can also be used to compute the derivative of a using the Chain Rule This leads to the following quotquoti rule and cor rule Theorem 1 Let a be a positive real number that not equal to one Then 7 a a lna 9 d1 and y i x al i 10 d 7 a J lna where C an arbitrary constant In your previous calculus cours 39ou learned the power rule for differentiation 0 n l 11 where n is any integer Using the definition of exponentiation we can write J em and use the Chain Rule to prove that the power rule actually holds for any real number n General Logarithmic Functions The natural logarithmic function can also be used to define the logarithmic function with an arbitrary base a Such a definition is necessary since certain bases such a 2 and a 10 are frequently used in application areas such computer architecture or numerical analysis To obtain a useful expression for the logarithm of a number J in base a we use the exponential function with base a De nition 2 General Logarithmic Function Let a be a positive real number The logarithmic function with base a denoted by loga de ned by the relation loga J y ltgt 17 J 12 where J a positive real number and y a real number The left side of equation 12 is known the logarithmic form while the right side is called the exponential form Often it is necessary to convert from one form to the other to solve an equation involving exponential or logarithmic functions Taking the natural logarithm of both sides of all J and rearranging algebraically we obtain a more useful form of the definition an lnaquot loga J 13 which is known the changeofbase formula This definition can easily be used to obtain the following differentiation rule and corresponding integration rule Theorem 2 Let a be a positive real number Then 1 1 i 0 7 14 11100 11 1lna and 11117 0 15 a W lna H l where C an arbitrary constant Example 2 Differentiate log2sin sinlog2 Solution Using the Product Rule the Chain Rule and the rule log 1 1 1 ln 1 we obtain f 1 sinlog21log2sinx39 log2sinxsinlog2139 1 1 g I n 7 mm 002 J sin1 ln 2 1 ln 2 cos1 logz sin 1 coslog2 1 lquot lquot 39lquoti sin ogz1cos1 ogzsin1cosogz1 16 sin ln 2 1 ln 2 Cl Example 3 Differentiate f 1 log3ex log21 Solution Using the Chain Rule we obtain 1 ex logz 1 ln 3 f39U l 63 logz 139 1 x 3 7 1 ex log21 ln3 1ln2 Just the natural exponential function is the inverse of the natural logarithmic function the exponential function with base a is the inverse of the logarithmic function with base a This rela tionship provides a useful tool for computing inverse functions the following example illustrates Example 4 Given f1 log21 32 find f 1 1 Solution Setting 9 f 1 1 and using the cancellation equation ff l 1 1 yields y 1 or log2y 32 18 Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 7 Notes These notes correspond to Section 61 in the text In the next few lectures we begin to expand the applicability of the definite integral In particular we will learn how the definite integral can be used to compute areas of more general regions than in previous discussion and even volumes of certain threedimensional solids Areas Between Curves We have already learned that the definite integral 4 m d1 1 can be used to compute the area of the region bounded by the curve 1 f 1 the horizontal line 9 0 and the vertical lines 39 a and 1 I Now suppose we need to compute the area of a somewhat more complicated region one that is bounded above by the curve 1 f 1 and bounded below by the curve 1 g1 between the vertical lines 139 a and 1 b We assume that f 1 and y1 are continuous on 11 and that f 1 2 y1 on 21 As before we can approximate this region using rectangles We divide the interval 11 into n subintervals of width A1 b an These subintervals are 1011 1112 17417 where 1 a iA1 for i 0 n We then approximate the region by n rectangles of width A1 and height where is any point in the ith subinterval 15415 Note that the height the ith rectangle happens to be the vertical distance between the curves 1 f 1 and t y1 at the point Using these rectangles we can approximate the area A of the region using the Riemann sum R A RR ELMf UV lAJ r 2 i1 As the number of rectangles n approaches infinity we obtain the exact area of the region between the curves 1 f 1 and t g1 which is given the definite integral A RR b M max 3 We can therefore use the definite integral to compute the area of the region bounded above and below by any two curves not just regions bounded above by the curve 1 f 1 and bounded below by the horizontal line 9 0 A similar approach can be used to compute the area of the region bounded on the right by a curve of the form J y on the left by a curve J gy below by the horizontal line 9 c and above by the horizontal line 1 d If y 2 gy on the interval 0 then the area A of such a region is given by the definite integral d A fy yydy 4 On a standard graph J increases to the right so in order to ensure that the area has the correct sign the function whose graph is the left boundary of the region in this case 99 is subtracted from the function whose graph is the right boundary which is y in this case For this reason it is helpful to remember right left when integrating with respect to y to compute the area between to curv Similarly it is helpful to remember top bottom when integrating with respect to J since the integral represents the correct area when the function whose graph is the bottom boundary is subtracted from the function whose graph is the top boundary The assumption that fJ 3 gJ or that y 3 gy should not be ignored In general the above integrals represent the net area of regions between two curves If gJ 2 fJ on any subinterval of 21 then on that subinterval the area between the curves is counted negative 7 toward the value of the integral of fJ gJ from a to b In general the area A of the region between the graphs of fJ and gJ from J a to J I is dJ 439 L41 V 4 A foo m where the absolute value ensures that the area between the two curves is always counted positively A similar integral can be used to compute the area between the curves J f y and J 19 from y c to t 1 Example 1 Use calculus to compute the area of the triangle with vertices 42 17 and 5 3 Solution The triangle is displayed in Figure 1 We will compute its area by computing the area enclosed by the three lines that define the edges of the triangle We begin by computing the equations of these lines First we consider the line that passes through the vertices 4 and 1 7 The slope is given by 7 2 5 7 1 6 1 4 which yields the equation y 71J 1 7 or J J 6 8 412 Triangle with vertices 42 17 and 53 X Figure 1 Triangle with vertices 42 17 and 5 3 Proceeding in the same manner with the other two lines we obtain slopes 3 7 10 5 3 2 73 7 77 7 7 9 1 4 239 4 9 39 gt and the corresponding equations y r x 1gt y 2 x 4gtgt 10gt which simplify to z 19 4 2 y 7JJ7 y 3J 7 11 2 2 9 9 We are now ready to compute the area of the triangle using definite integrals We divide the triangle along the dashed line shown in Figure 1 This yields two smaller triangles the area of which can be computed using a single definite integral in each case The left triangle is the region bounded by the lines J 1 y J 6 and t 5J9 29 The line 9 J 6 defines the top of the triangle while the line 1 2 defines the bottom The limits of integration are dictated by the right boundary J 1 and the fact that the two lines 9 J 6 and t 5J9 2 9 intersect at J 4 It follows that its area which we denote by Al is AI J396 dJ 1 V a 2 6 7 7d 74J 9J9 J 9 545 2d 7J 7 7J 7J 49 9 9 9 74 14 J2 1 7 4 39 21 350 18 175 12 9 We use the same approach to compute the area of the triangle to the right of the dashed line This triangle is the region bounded by the lines J 1 t 5J2 192 and t 5J9 29 The line 1 5J2 192 defines the top of the triangle while the line 1 5J9 2 9 defines the bottom The limits of integration are dictated by the right boundary J 1 and the fact that the two lines 1 J392 192 and t 5J9 29 intersect at J It follows that its area which we denote by Ag is JJ39ESJEU 1 2 2 9 9 J gxidx 1 18 18 18 18 175 0 EJ 18 dJ 1 7 4 139 181 JJJl JdJ 1 J2 7 37 17 18 12 NJ 1 M32 53 E Ju 1m1 1U 1 875 Jquot i F 7 1 18 2 81l 2 l1 1 840 E 700 1 420700 18 280 18 140 13 9 We conclude that the area A of the entire triangle is 175 140 315 AA Ali 775 14 1 2 9 9 9 J Cl Example 2 Compute the area of the region bounded by the curves 9 sinJ and y cos J well the lines J 0 and J 2r Solution The area that is to be computed is shown in Figure 2 The region whose area we will compute is shaded in the figure We can see that this region can be divided into three regions 0 the region bounded above by y cos J below by y sin J to the left by J 0 and to the right by 774 which is where y cosJ and y sinJ intersect o the region bounded above by y sinJ below by y cos J to the left by J 7r4 and to the right by J 5774 which is another point at which 9 cosJ and y sinJ intersect o the region bounded above by y cos J below by y sin J to the left by J 37r4 and to the right by J 2r The area of these regions can be obtained by evaluating the definite integrals 27 Al cosJ sinJ dJ AZ sinJ cosJ dJ Ag cosJ sinJdJ 15 l 0 f 7r l Using antidifferentiation rules we obtain cosJ sinJdJ sinJcosJ C 16 It follows from the Fundamental Theorem of Calculus that Al cosJ sinJdJ 0 A74 0 sin7r4 cos7r4 sinO cos 0 sinJ cosJ 01 xi 1 17 Area between ysinx and ycosx 0 S x S 21 15 1 Figure 2 Region bounded by y sin 1 y cos 1 J 0 and J 27 shaded AZ 2 sinJ cosJdJ t in cosJ s J M4 cos57r4 sin57r4 cos7r4 sin7r4 l gtll l 77 77 2 5 18 and o27r A cosJ sinJdJ sinJ sin27r cos27r sin57r4 cos57r4 xi xi 0 1 7 7 1 2 2 1 xi 19 We conclude that the area A of the entire region between the curves is given by A2211AZA 12 1 4 20 It should be noted that this area is given by the definite integral o27r cosJ sinJ dJ 21 i 0 which we computed by dividing the interval 0 2r into subintervals on which cosJ sinJ is either positive or negative but not both Example 3 Compute the area of the region bounded by the line 9 J and the parabola J 92 2 Solution This region is shown in Figure 3 The curve J 92 2 cannot be described by an equation of the form 9 fJ because it does not pass the vertical line test Therefore it is dif cult to compute the area of the shaded region in the figure by integrating with respect to Instead we will integrate with respect to y with our integrand being the distance between the Area between xy and xy22 Figure 3 Area between J y and J 92 2 shaded Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 20 Notes These notes correspond to Section 82 in the text Trigonometric Integrals So far we have obtained several integration rules by reversing various differentiation rules Addi tional integration rules can be obtained by employ 1g trigonometric identities to convert integrands involving trigonometric functions into a form in which other integration rules can be applied For such integrands the resulting rules can then be used directly instead of continuing to rely on the trigonometric identities used to establish them For example suppose we wish to evaluate an integral of the form sinm J cos J dJ 1 If n is odd then we can use the identity sin2 J cos2 139 1 to exprc n 1 of the powers of cosine in terms of sine Then we can use the substitution u sin Because du cosJdJ the resulting integral with respect to u will be simple to evaluate since the integrand will consist of several terms of the form umwk for k 0 n 1 2 which can easily be integrated using the power rule An integral of the form 1 where m is odd can be handled similarly using the same identity Example 1 Evaluate sin3 J cos2 J dJ 2 Solution Since the power of sine is odd we write all but one power of sine in terms of cosines U3 2quot 2 2 y sin Jcos J dJ sinJ1 cos J cos JdJ sinJ cos JdJ sinJ cos JdJ 5 Using the substitution u cos J with du sinJdJ yields sinJcos2JdJ sinJcos4JdJ tt2d1Ltt4dtt C If both m and n are even then the half angle formulas 1 i 1 sinz1 1 cos 21 cos21 1 cos 21 A m V can be used to reduce the powers of sine and cosine until every term in the transformed integrand either has an odd power of sine or cosine or until a simpler integration rule can be used If a one may find it easier to use the doubleangle formula 1 sin1 cos1 i sin 21 Example 2 Evaluate sing 1 cosz 1 11 Solution Using the doubleangle formula sin 21 2 sin 1 cos 1 2 g 2 sinz1cos21d1 11 We can then use a halfangle formula yields 1 2 7 2d 41 sin 1 1 2 1 cos21 sin 1 2 toobtain 1 1 m 4 sinz1cos21d1 il dx 4 2 1 7 1 cos4111 8 1 sin41 7 x 7 8 4 1 sin 41 8 32 Cl Integrals of the form tan39 1 sec 1 d1 2 term of the form sin1cos1 arises this can be handled using the substitution u cos1 or u sin1 but 6 11 1 can be handled using a similar strategy In this case the useful identity is sech 1 tanZJ If the power of secant is even then all but two of the powers of secant can be expressed in terms of tanJ using this identity and then the substitution u tanJ can be used to obtain a simpler integrand If on the other hand the power of tangent is odd and n gt 0 then all but one of the powers of tangent can be expressed in terms of sec J and then the substitution u seCJ can be used since we would have du seCJ tan Example 3 Evaluate S C4JEan3JdJ 13 Solution We write all but two powers of secant powers of tangents using the identity sech tanZJ 1 14 This yields 43 t22 3 25 23 r seL Jtan JdJ lsen Jtan J1tan JdJ seL Jtan JdJ lsen Jtan JdJ L1 Using the substitution u tan J with du sech1J yields c4 J tan3 J dJ sec2 J 611101 dJ sec2 J tan3 J dJ u5 du u3du 3 4 U U i c e 4 3 4 tan 7 tan 7 7 a 16 6 4 An alternative approach is to write all but one power of tangent powers of secants using the identity tanZJsech 1 17 This yields sec J tan3 J dJ sec Jsec2 J 1 tan J dJ sec J tanJ dJ sec4 J tan J dJ 18 Using the substitution u sec J with du secJtanJ dJ sec4Jtan3JdJ sec6JtanJdJ S C4JtanJdJ uodu u3du 3 J sec J tan J dJ JsecJtanJdJ 19 Other cases are not straightforward It may be necessary to use other trigonometric identities integration by parts or the rules for integrating tanJ and sec J C 20 tan J dJ ln secJ C 21 cJ1J lnsecJ tanJ Example 4 Evaluate sec3 J tan2 J dJ 22 Solution Since the power of tangent is even and the power of secant is odd it is not possible to rewrite powers of one in terms of powers of the other and use a simple substitution in the previous example Instead we can proceed using integration by parts with u secJtan2 J 11 secZJ 23 which implies du 2 sec3 JtanJ secJtan3J dJ r tanJ 24 Before applying integration by parts we use the identity tanZJ SQCZJ 1 to write du 2 s gJtanJ secJsec2J 1 tan J dJ 2r c JtanJ C2 7 80C 7 tan 7 80C 7 tan 3 se gJtanJ secJtanJdJ 25 Integrating by parts we obtain t32 3 c3m2 sec Jtan JdJ secJ tan J J sec Jtan JdJ secJtan J1J Rearranging algebraically yields l32 3 4sec Jtan JdJ secJtan J secJtan JdJ or 9 1 1 9 sec3 J tan2 J dJ 1 secJ tan3 J 1 secJ tan2 J dJ We now focus on the remaining integral 2 sec J tan J dJ Rewriting the powers of tangent powers of secant yields secJtanZJ1J S CJS CZJ 1 dJ sec3 J secJ1J To integrate SQCJ we proceed follows using the substitution u secJtanJ sec J dJ secJ tanJ 2 secJ tanJ secJ dJ secJ s secJtanJ secJ tanJ u ln M C ln secJ tanJ Cl 39e use integration by parts with To integrate sec3J u sec J 11 SQCZJ dJ which yields du secJtanJ dJ r tan It follows that y y sec3 J dJ secJ tanJ secJ tan2 J dJ secJ tanJ with du 26 29 30 31 3 33 34 Using the identity tanz1 sec21 1 yields sec3111 sec1tan1 sec1sec21 1 11 sec1tan1 sec3111 sec111 35 Rearranging algebraically yields Esec3111 sec1tan1 sec111 36 I 1 1 sec3111 sec1tan1 i sec111 37 Applying our earlier result of integrating sec 1 we obtain 1 Finally we put all of the pieces together and obtain 1 1 sec1tan1 lnsec1tan1 38 1 1 806 J tang J 1 80C 611131 1 80C tang J 1 I 1 I asecxtan31 a sec 111 1 3 1 1 1 asecxtairxFE sec1tan1 lnsec1tanx lnsec1tanx 1 3 1 1 1 asecxtan 1 a sec1tan1 7 lnsec1 tan1 2 1 3 1 1 f a sec1tan 1 g sec1 tan1 g ln sec1 tan1 39 Cl Trigonornetric identities can also be helpful for integrals of the fornls sin m1 cos 711 11 40 sin m1 sin 111 11 41 or cos m1 cos 711 11 42

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