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# Calculus I MATH 1210

Utah State University

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This 56 page Class Notes was uploaded by Benton Yundt on Wednesday October 28, 2015. The Class Notes belongs to MATH 1210 at Utah State University taught by James Cangelosi in Fall. Since its upload, it has received 47 views. For similar materials see /class/230416/math-1210-utah-state-university in Mathematics (M) at Utah State University.

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Calculus Refresher Course Fall 2007 Review of Calculus I and Calculus II Material Prepared for the Fall 2006 AlgebraCalculus Review Workshop for engineering majors who have had a gap of one year or more in their study of mathematics I I PreviewOverview 0f Calculus The calculus sequence at Utah State University consists of three courses 1 Math 1210 Calculus I 2 Math 1220 Calculus II and 3 Math 2210 Multivariable Calculus All three courses are taught from the same textbook Calculus Concepts and Contexts 3rd Ed by James Stewart Honors sections may require different andor additional texts Calculus I and the first part of Calculus II focus on the two main branches of calculus ali erential calculus and integral calculus as they pertain to realvalued functions of a single variable and to a limited degree parametric equations In your own time you may wish to peruse the section in the textbook titled A Preview of Calculus that precedes Chapter 1 for a simple short overview of calculus and its branches The study of sequences and series is regarded by some as a third branch of calculus These topics found in Chapter 8 of the textbook are studied in Calculus II The Calculus II course also includes the extension of differential and integral calculus l to functions other than realvalued functions of a single variable such as vectorvalued functions and 2 to coordinate systems other than the rectangularCartesian coordinate system such as the polar coordinate The Multivariable Calculus course is a continuation of the extension of differential and integral calculus of functions of a single variable 1 to realvalued functions of several variables 2 vector fields and 3 to the cylindrical and spherical coordinate systems in threespace The foundation of all branches of calculus or the common thread that ties all branches together into one discipline called calculus is the notion of limits Differential calculus is founded upon the following definition for the derivative function f 39 of a function f f 39x lhinol This is called a di erence quotient fx h fx h Question In terms of the graph of f what does these represent Integral calculus is founded upon the following definition for the de nite integral 17 f x alx of a function f 17 n if L fxd as f x Ax This is called a v Riemamm Note The key structure of the Riemann sum is a product of f times a small change in x The sum of these products has widespread applications 1 Calculus Refresher Course Fall 2007 Besides the fact that derivatives and de nite integrals are both de ned in terms of limits we will review another remarkable connection between these two branches of calculus established by The Fundamental Theorem 0f Calculus The de nition of a convergent sequence and the de nition of a convergent series are also based on limits We will spend little if any time reviewing this Calculus II material since it is not pertinent to the Multivariable Calculus material studied in Math 2210 I II The De nition of the Derivative f 39 of a function f Recall f 39x One can see the derivative of a function f is de ned in terms of the limit of the quotient This quotient is called a difference quotient This quotient without the limit is the slope of a secant line containing the points x f x and x h f x h on the graph off It represents an average rate of change in the function f between these two points The interpretationmeaning of a difference quotient and subsequently of its limit the derivative f 39 depends on the contextmeaning of the function f Consider the following three examples of various possible contextsmeanings of the function f Example 1 Suppose f x gives the distance ft of an object moving in a straight line from some reference point as a function of time sec In this context the numerator f x h f x of the difference quotient represents the change in distance of the object in the time period beginning at x seconds and ending at x h seconds The denominator h of the difference quotient represents the time elapsed or change in time for the same time period Therefore in this context the difference quotient f x h f x h represents the average velocity ft per sec of the object over the time period beginning at x seconds and ending at x h seconds Example 2 f x gives the distance mi a car travels from some reference point as a function of the amount gal of gasoline consumed In this context the numerator f x h f x of the difference quotient represents the distance of the car travels from the time x gallons of gas have been consumed to the time x h gallons of gas have been consumed The denominator h of the difference quotient represents the amount of gasoline used during the same time period Therefore in this h context the d1fference quot1ent W represents the average fuel economy ml per gal of the car from the time of x gallons of gas consumption to the time of x h gallons of gas consumption Caleulus Refresher Course Fall 2007 Example 3 x ls merely regarded as the yrcoordmate parred wrth x of a pornt on the graph of eeantllne V xhx f xhx 1n thrs eontent the nurnerator xhr x othe dlfference quotrent represents the vemcal ehange or nsequot from the pornt xx to the pornt xhxh on the graph of The denornrnator h othe dlfference quotrent represents the honzontal ehange or runquot from the pornt xx to the pornt xhxh Therefore m thrs eontentthe 4 pl H d Fr m the pornt xx to the pornt xhxh on the graph of r gal fuel eeonorny ml pergal slope nsequot per runquot There are many other eontents thatwe uld n ld r X h h rto xh appronrrnates the lns antaneous rate of ehange at xquot as the ehange m the xrcoordlnate from rto xh r e h gets closerto zero 39T h r F r 39x lhmgw Aeeordngly lffls a dstanee funeuon of urne then ls the s 39 x glves the lnstantaneous slopequot ofthe graph ofat the pornt xx The lln specl c pornt aa havlng the same slope as the graph offatthls pornt that rst e through a he llne through aa whose slope ls glven by f39a ls ealled alangent 1mg or more specl cally Llnear Drstanee Tlme m seeonds Average veloclty Instantaneous Veloclty Arnount ofgas used GeneneFuneuon Coordlnate Slope ofseeantllne Calculus Refresher Course Fall 2007 III Limits and Continuity Since all branches of calculus are based on the notion of a limit an understanding of limits is necessary to understand calculus well In particular the derivative f 39 of a function f is def39med as the limit of a difference quotient m lim fx h fx hgt0 For any given function f and any fixed value of x in the domain off the value of the difference f quotient w depends only on the value of h that is the difference quotient is a real Valued function of the single variable h Since derivatives are defined as limits of difference quotients and since for each fixed value of x difference quotients are realvalued functions of a single variable to understand derivatives well one must first understand limits of realvalued functions of a single variable well Consider a realvalued function f of a single variable x Intuitively but loosely speaking if the value of f x approaches a real number L as x approaches a we say the limit of f x as x approaches a is equal to L denoted by lim f x L That is restricting the values of x near a results in function values that are tending toward a single value A key point to be made is that in the limit process we NEVER consider the value at x a In essence we mean lim f x x a We can examine limits numerically algebraically and graphically For illustration we begin with a graphical approach Examine the three similar but different functions illustrated below WE gJgkalv r39 i at 5 a Note a is not in the Note a is in the Note a is in the domain of f ie fa domain off but a domain off and fa does not exist but is not equal to L is equal to L values arbitrarily close to a are in the domain off In each case the value of f x approaches L as x approaches a ie lim f x L in all three examples However in the first example f is not defined at a in the second example f is defined at a but lim fx L at fa and in the third example lim fx L fa Calculus Refresher Course Fall 2007 The previous examples illustrate that lim f x may exist whether or not a is in the domain off Furthermore ifa is in the domain off lim fx may or may not be equal to fa Contrast the functions illustrated above to the functions illustrated below 3 In each case of the illustrations directly above the value of f x does not approach a single specific real number as x approaches a In cases such as these we say lim fx does not exist Ha because restricting the values of x near a does not result in function values that are tending toward a single value As for the cases where lim f x does exist the existence andor value of f a is irrelevant in the case where lim f x does not exist The concept of a continuous function is inextricably connected to the concept of the limit of a function A function f is said to be continuous atwhere x a provided lim f x L f a as is the case for the third function of the first three illustrated functions That is the limit and the function value at a are equal Intuitively a function is continuous atwhere x a if a small change in the domain represents a small change in the range ie x near a implies fx is near fa A function is continuous on an open interval cd provided it is continuous at all points in that interval Intuitively a function is continuous on an open interval cd if the graph of the function can be traced on that interval without lifting one s pencil When it comes to evaluating limits which is crucial to understanding derivatives it is helpful to be able to determine whether or not a function is continuous Based on the intuitive descriptions of continuity given above and our familiarity with typical polynomial rational trig inverse trig exponential logarithmic and radical functions we can conclude that all functions in this list as well as combinations of these functions by addition subtraction multiplication division and composition are continuous where they are de ned Of course no function is continuous where it is unde ned Accordingly if f is one of the functions described above and if a is in the domain off then lim f x exists and this limit value can be found by direct substitution ie lim f x can be found by evaluating f a because by definition continuity means lim f x f a Calculus Refresher Course Fall 2007 Consider the following examples Example 1 Discuss the continuity and limit of the given function at the give value for a Z fxsin x 61 Since f is a composite function consisting of a trig function composed with a radical function both are continuous over their domains we conclude that f is continuous where it is defined Since the given value of a is in the domain off we conclude that f is Z Z 2 continuous at a S1ncef1s continuous at g the 11m1t as x approaches can be found by direct substitution as follows 2 2 lim sinJf L sinquotisin i Hi 9 9 3 2 9 LA Example 2 Discuss the continuity and limit of the given function at the give value for a 7 fx e392m cos32 x a 3 Since f is a combination of exponential trig and radical functions we conclude f is continuous where it is defined Unfortunately the given value for a is not in the domain of f Since the given value for a is not in the domain off we conclude that f is not continuous at a However the existence of lim f x is independent of the existence of f a Thus we must further analyze f to determine whether of not lim f x ex1sts Intu1t1vely as x approaches 3 tan x gets 1nf1n1tely far from zero and cosy2 x an X an X approaches 0 Thus 6 cos32 x gets infinitely far from zero ie e cos32 x does not approach a single specific real number as x approaches This may be verified numerically and graphically by examining the function values near 3 Since 32 7239 atm cos x does not approach a Single speci c real number as x approaches 3 we an x conclude 11m 6 cos3 2 x does not ex1st Hg Comment Graphical examination of a function allows us to visually inspect the function values near x a Numerical examination would involve literally evaluating the function at many x values near a However you cannot prove a limit exists graphically 0r numerically Calculus Refresher Course Fall 2007 Example 3 Algebraic Examination Manipulation Discuss the continuity and limit of the given function at the give value for a x2 9 x 3 a3 fx As in the previous example the given function is not continuous at a because it is not 2 x defined at a However further analys1s 1s requ1red to determine whether 11m Xgt3 9 ex1sts x 3 or not Now 2 Q W is equal to x3 for all values ofx except 3 Since x 3 x 3 the value of f at 3 is irrelevant when considering lim f x we conclude Xgt3 2 lim x 39 lim x3 Since all polynomial functions are defined and continuous for all xgt3 x xgt3 x2 9 values ofx we conclude ling 3 lin1x3336 H x H The moral of the story of Example 3 is twofold 1 Even though a function f may be undefined atwhere x a lim f x may exist and 2 manipulating f algebraically without changing the values of x except possibly at x a may lead to a limit that can be evaluated by direct substitution One is well to remember this lesson when evaluating limits that result from applying the definition of a derivative ie f 39x IhingW After all regardless of which function f one uses the difference quotient is always undefined at h 0 fxhfx h Accordingly to evaluate l im requires some sort of algebraic manipulation gt0 fxhfx h kwith the goal of quot 39 quot h as a factor of the 39 39 l Consider the following examples Example 4 Find the derivative function f 39 of the given function f fxx3 3 3 3 2 2 3 3 f39x mfxh fxlimxh x zlimx 3x h3xh h x hgt0 h hgt0 h h 0 h 1im3x2h3xhz h37 f zlim h3x2 3xhh2 zlim r3x2 3xhh2 3x hquot h heo h hgt0 Calculus Refresher Course Fall 2007 In the last step the limit was found by direct substitution of h with 0 due to the fact that our goal of eliminating h from the denominator had been accomplished leaving a function that is continuous at h 0 Another notation for f 39x is i f x Using this style of notation we have just shown if 3x2 Example 5 Find the derivative function f 39 of the given function f fx W fxhfx 1imJxh5 4 h fx1 3 M h quotxh5 xm Jxh5xm h Jxh5 lim hgt0 xh 5 x5 zlim xh5 x 5 i lt gt ieo Jxh5m M hxh5 m 11 hXgt li mt h39iIolh xh5m 2amp3 Xxh5m 1 l l Jxhgt5xlx5 Vx5xlx5 2Jx5 d 1 Therefore ixlx 5 7 dx 2V x 5 Example 5 demonstrates the technique of manipulating difference quotients involving radicals by multiplying both the numerator and the denominator by the conjugate of the numerator Calculus Refresher Course Fall 2007 Example 6 Find the derivative function f 39 of the given function f 1 fx a 1 i f39x Qcmm 3xh 3x 3xh 3x 1m 11m M h3xx 11 H WNW MM xxh 11m 11m Ho hlt3xx 11 hgt0 hlt3xxh l 1im lim limil 72 mo h3xx h mo fr3xx h mo 3xx h 3x Therefore i dx 3x 3x2 This example demonstrates the technique of manipulating difference quotients involving compound fractions by multiplying both the main numerator and the main denominator by the lowest common denominator of the secondary denominators Take some time to respond to the following items 15 Determine whether the given function is continuous at the given value of a Explain your reasoning I fxsecx 61 2 fxjsecx 612 Calculus Refresher Course Fall 2007 3 fx 1 a0 x 2 lnx 4 fx aZ x 2 lnx 5 fx ae x 2 69 Evaluate the given limits 6 Emmi H2 x2 gt1 lit 2 cos x 7 Hi 4 8 lim 1 Consider the behavior from both sides of zero Ho 1 87A Calculus Refresher Course Fall 2007 9 The limit obtained by applying the de nition of derivative to the given function fxx2 4x 10 The limit obtained by applying the de nition of derivative to the given function fx i x3 11 The limit obtained by applying the definition of derivative to the given function fx J Calculus Refresher Course Fall 2007 IV Differentiation Rules and Derivatives of Basic Functions Just as we were able to use the de nition of derivative to establish dx 2 ii etc The formulas on reference page 3 in the back cover of the textbook please nd a copy of this reference page on p 13 of this document were also derivedproved using the fxhh fxBy Iquot familiar de nition f 39x lim these formulas one is 40 able to compute the derivatives of various sorts of functions without appealing directly to the limitde nition for derivative The formulas used most frequently in the Calculus ICalculus II Multivariable Calculus sequence are enclosed in boxes but you may encounteruse some of the other formulas sporadically Formula 1 F1 for short states the derivative of any constant function is 0 Formulas F2 F3 and F4 state that the derivative of a constant times a function is the constant times the derivative of the function the derivative of a sum is the sum of the derivatives and the derivative of a difference is the difference of the derivatives respectively The Power Rule F8 shows how to differentiate ie nd the derivative of the power function f x xquot for any realnumber d exponent n Not1ce that our result 6799 3x2 from example 4 on page 8 1s 1ndeed cons1stent x with the Power Rule Combining these ve formulas F1 F2 F3 F4 F8 one can differentiate a variety of functions including all polynomials Example 1 Find the derivative of the given function fx 6x5 3x4 12x2 5 I d 5 4 2 d 5 d 4 d 2 d f x 6x 3x 12x 5 alt6x3x E12x5 d 5 d 4 d 2 4 3 1 4 6 7x 3 7x 12 7x 0 65x 34x 122x 30x 1293 24x dx dx dx Therefore f39x 30x4 1293 24x With a little practice every calculus student should be able to differentiate any polynomial almost automatically without going through all the intermediary steps shown above The next example is similar with some of the steps omitted Example 2 Find the derivative of the given function fx x7 8x3 le f39x i0 8x3 12x 7x6 83x2 12lx0 7x6 24x2 12 Calculus Refresher Course Fall 2007 EDIFFERENTIATIONRULES Si l hs d d z 7 I dx E 0 m6 0 if A I r d v 3 ELM 906 f x 90 4 ENG 50 f x g x Ii i wg 5 dquot fq2 fg A gf A Product Rule 6 Ix 9 9 Quonent Rule 2 I I 7 IfIngD f gxg39x Chain Rule 8 i 1quot nx39H Power Rule v Ix IX I I EXF ONENIIAL LOGARETHMIC FUNCTIONS I r 9 iIn e IO iIaquota lnu 1L dquot I 1 d l lIZ111Al 12 dXIloan7X1nu quotTiileowom ac FUNCVTIONS 39 I I I r 13 ilsin x cos I I4 icosx sinx 15 i lanx seclx I 39 r 7 IA dx l6 71csc1 cscx cotx l7 1amp6ch secx um 18 1 col 39 csc3x zl39 Ix Ix y me INVERSEVTRIGIONor ETRIE FUNCTIONS WW d 1 l8 4 7 ll5m I 1 I 39 1 quot71 m 39 quot1 223 39 23 quot7 24 1 IX 68 A x 3 i I Ix m X 1 x3 1 IV L01 0 1 x1 HYPERBOLIC FUNCTIONS I 25 Lsinhx coshx 26 If coshx sinhx 27 L lanh x sechlx dx l39 d39 I 28 cschx csch x cothx 29 i sechx scchx Ianhx 30 I cuthx cschlx Ix A LY iNVEIISE HYPERBC LIC FUNCTIONS I l d 1 I l 3L 391quot 32 hquot 33 quot ix sun 1 31 LY cos x x 1 IX nh l i x2 d 1 I l d 39 34 quot5 7 35 l lquot 36 quotg 1 JV csch dx sec 1 Ix coll l VA 393 LA Calculus Refresher Course Fall 2007 Example 3 Find the derivative of the given function fx6x x l4 x f39xi 6x x l4 i6xx 7x394 dx x dx 2 12 l 28 3 1 y 7 76x 7x 2 4 7xsjiii 5 2 5x 2J5 x5 The lesson learned from this example is that sometimes one can manipulate a function rst making it easier to differentiate Ask yourself which is easier to do rst 7 algebra or calculus The answer will vary from problem to problem The following example adds formulas F13 F14 F15 and F17 into the mix Example 4 Find the derivative of the given function tanx fx 6secx 3cosxs1nx 2 sec x f39x 6secxtanx 3 s1nxcosx 7 Clearly the derivative of any polynomial function is a polynomial function The derivative of a trig function is a trig lnction or a combination of trig functions The derivative of the exponential function is again the exponential function F9 Formula F11 is rather bold in the sense that it breaks the mold we just established formula F11 states the derivative of the natural log function is not a logarithm or combination of logarithm functions but a different type of function called a rational function ie quotient of polynomials Example 5 adds formulas F9 and F11 to the mix Example 5 Find the derivative of the given function fx3lnx l3e cosx x f39x i 3ln x 13 co x s 2i2 3e sin x recall the second term 1 can dx x x x x be rewritten as x 1 and differentiated using the Power Rule Calculus Refresher Course Fall 2007 Recall formulas F3 and F4 establish that the derivative of a sum is the sum of the derivatives and the derivative of a difference is the difference of the derivatives It is natural for beginning calculus students to make the erroneous assumption that the derivative of a product is the product of the derivatives and the derivative of a quotient is the quotient of the derivatives Consider the functions f x 2x and gx E to quickly debunk such a wild yet understandably natural x guess d d d 3936 72x2 Clearl then 39x 2 2 7x 20120 f d c y ful ll L g39OC 13 1 Clearly then g39x 7 L2 9 0 dxx x2 Tx 1 Although more complicated than simply multiplyingdividing derivatives there are formulas called the Product Rule and the Quotient Rule F5 F6 for differentiating products and quotients of functions Similarly there is a formula called the Chain Rule F7 for finding derivatives of composite functions The following examples add these three formulas into the miX Example 6 Find the derivative of the given function fx lnxtanx tanx In xsec2 x f39x ilnx tanxlnx itanx ltanxlnxsec2x dx dx x Example 7 Find the derivative of the given function x2 2xl x 6 d 2 2 d ape 2xlx 6 x 2xlax 62x2x6x22x11 HY Mr 2x2 12x 2x12 x22x lx2 12xllx llx l ye 62 ye 62 ye 62 fx f 39x Calculus Refresher Course Fall 2007 Example 8 Find the derivative of the given function hx I 3 4x The given function hx is a composite function f gx where f x J and g x x3 4x By the Chain Rule 3x2 4 1 hr t 39 3 2 4 x fgx g x 2 x34x x 2 x34x Example 9 Find the derivative of the given function hx xsin x As in the previous example hx is a composite function f gx Try to identify the inner function g and the outer function f and check the accuracy of the following computations cos x l h39 x 7cosx 2Vsinx ZJsinx Example 10 Find the derivative of the given function hx sin J Here hx is composed of the same two functions as in the previous example However the roles of the inner and outer functions have been reversed cos J 2J5 Example 11 Find the derivative of the given function x In x x f x 5 The Quotient Rule is required to differentiate the given function However there is a product of functions in the numerator which will require the Product Rule to differentiate In other words to differentiate f requires the Quotient Rule with a Product Rule nested within h39x cos results of the Product Rule applied to the numeratorx2 5 x In x2x 962 52 f 39x llnxxix2 5 xln x2x lnx1x2 5 xln xxzx x2 52 x2 52 lenx Slnxx2 5 2lenx xZ lenx Slnx S x2 52 x2 52 Calculus Refresher Course Fall 2007 Example 12 Find the derivative of the given function fx sin x em To differentiate this function requires the Chain Rule nested within the Product Rule f 39x cos x em sin x results of the Chain Rule applied to the second factor of the product cos x em sin x ems sin x em cos x sin2 x In the next example one must not only find the derivative of a function but do something with the derivative function once its been found The problem is to find the equation for a tangent line Recall from algebra that to find the equation for a line it is sufficient to know a point on the line and the slope of the line Recall also that the slope of a line tangent to the graph of a function is found using the derivative of the function Example 13 Find the equation of the line tangent to the given function atwhere x a fxtanx 61 The ycoordinate of the point of tangency is found by evaluating f at a Since f l the point of tangency is EA The slope m of the tangent line is found by evaluating 2 the derivative f39x sec2 x at a Therefore m sec2 2 Now that we have the slope of the desired tangent line and a point on it we use the pointslope formula and simplify to obtain the equation for the tangent line yy0 mxx0 7T l2 x 7 y l 4 7r l2x i y 2 y2xl i 2x7 7 y 2 2 y2x7 Caleulus Refresher Course Fall 2007 We have the followlng lllustxauon x tam zen y 2xi 2 The gure shows the graph of the pomt oftangeney andthe tangent hne 1 1 4 F amnlnlA E A L Fa x slnee m 1 the pomt oftangency ls 10 To ndthe slope othe tangent llne we 39 Then we mustrememberto evaluate ata To nd requlres a am Rule nested Wlthln a Quotient Rule 1 227 We are only lnterestedm 391 We could slmpllfy and then plug 1m forx but lt oplug must rst nd ch x slmpllfyithls ls avaluable lesson to leam 1 2 7 72 1 17 271 1 fag 242 1 J Calculus Refresher Course Fall 2007 Now we use the point 11 and the slope m 2 in the pointslope formula to get the equation for the tangent line y 1 2x 1 y l 2x2 y 2x3 Take some time to respond to the following items 110 Find the derivative f 39 of the given function f l fxx5 3x2 6x4 2 fx3e tanx secx7lnx 3 fxexlsx2 4 fx e39x 5 fxxE 6 fx1nJ Calculus Refresher Course Fall 2007 7 fxln ml 3xsin x 8 fx ex 9 cosxjx4 5x2 1 10 ex ux 1 l 12 Find the equation for the line tangent to the graph of the given function atwhere x a ll fxx2 x l a2 x2 ial x e 12 fx 20 Calculus Refresher Course Fall 2007 v Suppose we wrsh to ndthe hne tangenttothe eurpse gwen by 2 x2 1 at the pornt 2 An rssue to eonsrder 15 that there rs no funetron whose graph rs the entare e1hpsethe eurpse does not sausfy the vertrea1 hne testquot However the eurpse does pass J5 the vemeal hne testquot loeany nearthe pornt In faet the entrre upperhalf ofthe eurpse passes the vemcal hne test Indeed the graph of fx 942x2 obtatnedby solvrng the equataon othe eurpse fory and usmg the posture root rs the upper half of the elhpse Now we ean nd the equataon for the tangent hne usmg the rnethod revrewedrn the prevrous seetron We wru needthe denvatave off 79 1 x 39 7 718 7 x 24979x2 X J979x2 The slope othe tangent hne rs gwen by m 7M 2 2 x2 1 funetron ofx r e yx7at 1east1oea11y near eertatn pornts 1n the example wrth the eurpse p 39 39 Calculus Refresher Course Fall 2007 However it is not always possible to explicitly solve an equation that implicitly de nes y as a function of x for y Indeed the equation ey x2 y sin y cannot be solved for y Nevertheless this equation does implicitly de ne y as a function of x Fortunately there is a method referred to as implicit di rentiation that allows us to find the derivative y39 of this implicitly defined function Z We will first use implicit differentiation to find y39 where x2 y l Then we will use implicit differentiation to find y39 where ey x2 y sin y In these two cases and whenever we use implicit differentiation we will differentiate both sides of the equation with respect to x remembering to treat y not as a constant not as we would treat x but as a function of x Doing so will result in an equation that can be solved for y39 2 Example 1 The equation x2 y l implicitly defines y as a function of x Use implicit differentiation to find y39 First we differentiate both sides of the given equation with respect to x 2 952 if 1 Since y is a function of x y2 is a composite function y is the inside function Thus to differentiate the second term on the d xz L 11 lefthand side requires the Chain Rule dx 9 dx 2x 27y y39 0 9 Now we have an equation 2y y 2x we can solve for y39 y39 2x i 2y y 9x Note how the two results fix 1 718x compare 2 979x2 W That is because the explicit form was y fx 19 9x2 97 Notice the value of y39 at J33 2 can now be found as follows y39 T3 This is the same value we obtained with considerably more effort on the previous page 22 Calculus Refresher Course Fall 2007 Examplez The equation ey x2 y sin y implicitly de nes y as a function of x Use implicit differentiation to find y39 ey x2ysiny d y 2 d Recall y is treated like a function of x7 ER x y altsmy because it is precisely that locally y 2 I I e y 20 x y y cos y To differentiate the first term of the left eyy39 2xy xzy39 y39cos y hand side requires the Chain Rule To y 39 2 39 t 2 differentiate the second term on the left e39y x i y COS xy hand side requires the Product Rule To y 9y x 005 y 295 differentiate the only term on the right 2 9 hand side requires the Chain Rule y39 ey x2 cos y Take some time to respond to the following items 12 The given equations implicitly define y as a function of x Use implicit differentiation to find y39 1x2ylny 2 y3 secx ye 3 Find the equation for the line tangent to the graph of 2x2 y2 2 25x2 yz at the point 31 23 Calculus Refresher Course Fall 2007 VI Applications of Derivatives Recently we have reviewed how to compute derivatives of functions Let s not loose sight of the fact that all of the rules and formulas we have been using to differentiate functions are founded upon the de nition f 39x As noted when this de nition was first introduced the difference quotient represents various sorts of average rates of fxhfx h change depending on the contextmeaning of f and the derivative f 39 represents instantaneous rate of change Recognizing that f 39 represents instantaneous rate of change opens ones eyes to the many applications of derivatives to all fields of science business etc Time constraints prevent us from delving deeply into these applications No doubt you will encounter various applications of derivatives in future math classes and more importantly for you in courses within your major Recall that one interpretationapplication of derivatives the geometric one is that f 39x gives the instantaneous slope of the graph of f at the point x f x This idea can be exploited to garner lots of information about the shape of the graph off For example if f 39 is positive on an interval then the slope of f is positive on that interval and so the graph of f is increasing ie rises as one moves left to right on that interval Since the derivative f 39 of a function f is itself a function it can be differentiated to get what is called the second derivative f off eg fx 3x4 6x3 3 f39x12x3 18x2 3 f quotx 36x2 36x As does f 39 the second derivative f also provides information about the shape of the graph off f quot is positive on an interval 3 f 39 is increasing on the interval 3 the slope of the graph of f is increasing on the interval 3 the shape of the graph of f is quotconcave upwardquot This and other geometric applications of derivatives are summarized below First Derivative Test IncreasingDecreasing Test If f 39x gt 0 on an interval then f is increasing on that interval If f 39x lt 0 on an interval then f is decreasing on that interval Second Derivative Test Concavity Test If f quotx gt 0 on an interval then the graph of f is concave upward on that interval If f quotx lt 0 on an interval then the graph of f is concave downward on that interval 24 Calculus Refresher Course Fall 2007 The First Derivative Test for Relative Maximum and Minimum Values Suppose f is continuous atwhere x c and also suppose that c is a critical point ie f39c 0 or f39 is not differentiable at c This test examines if a function changes increasing to decreasing or decreasing to increasing at a point on its graph to identify local maximum and minimum values 0 If f 39 changes from positive to negative at c then f c is a relative maximum 0 If f 39 changes from negative to positive at c then f c is a relative minimum 0 If the sign of f 39 does not change at c then f c is neither a relative maximum nor a relative minimum Example 1 Find open intervals on which the graph of the given function f is increasingdecreasing also nd open intervals on which the graph of f is concave upwarddownward also nd all relative maximumsminimums fx x4 4x3 First we extract what information we are able from analyzing the derivative f 39x 4965 12x2 To nd critical points we solve the following equation 04xS 12x2 04x2x 3 0x2 0x 3 x0 x3 Therefore 0 and 3 are critical points Since f is a polynomial it is differentiable for all x thus there are no other critical points than those found above By testing points in the intervals determined by the critical points we make the following conclusions Since f does not change from increasing to decreasing or viceversa at 0 we conclude f0 is not a relative maximumminimum Since f is continuous at 3 and since f changes from decreasing to increasing at 3 we conclude by the First Derivative Test f3 which evaluated is 727 is a relative minimum 25 Calculus Refresher Course Fall 2007 Next we extract what information we are able from analyzing the second derivative f c12x2 7 24x Since f x exists for all x the concavity ofthe graph offcould only change from concave upward to concave downward or viceversa at values of x for which f x 0 0 12x2 7 24x 0 l2xx 7 2 0 x 0 x 7 2 x 0 x 2 By testing points in the intervals determined by these values we make the following conc us1ons Interval f quotx Conclusions about f Eh go ggzvii c g Iii121136 gt0 P05 concave upward where concavity changes is called an 0 2 neg concave downward in ection POW 501mg as 19 graph Off 2 00 p05 concave upward has a tangent line there Therefore 0f0 and 2f2 are in ection points Combining iitfumiatiunie aidin 39 39 39 concavity etc allows us to draw a reasonably accurate ske ch of the graph of f see gure 1 727 is arelative minimum The Second Derivative Test for Relative Maximum and Minimum Values Suppose f is continuous near x c and also suppose f 39c 0 o If f c gt 0 then fc is a relative minimum 0 If f c lt 0 then fc is a relative maximum 0 If f quot0 0 then the Second Derivative Test is inconclusive Calculus Refresher Course Fall 2007 Notice in the previous example f 393 0 and f quot3 gt 0 con rming with the Second Derivative Test this time around f 3 is a relative minimum The Extreme Value Theorem If f is continuous on 61 b then f has an absolute maximum and an absolute minimum on a b Note In this case the absolute maximum occurs at an endpoint ie a or b or at a critical point between a and b Similarly the absolute minimum occurs at an endpoint or at a critical point Example 2 Find the absolute maximum and the absolute minimum of the given function on the given interval fx 3x4 16x3 18x2 1 4 First a comment is in order The wording of the prompt for the task infers the given function f has both an absolute maximum and an absolute minimum on the given interval We know this is indeed the case because The Extreme Value Theorem guarantees it The Extreme Value Theorem can be invoked since the function is continuous on the given closed interval We know the absolute maximum and the absolute minimum each will occur either at an endpoint ie 1 or 4 or at a critical point Accordingly we merely have to identify these points and evaluate the function f at each of them to discover the absolute maximum and the absolute minimum We DO NOT have to determine which if any of the critical points correspond to a relative maximumminimum using the First and or Second Derivative Tests We will find the critical points evaluate the function at each critical point within the given interval and at the endpoints of the interval This is all we need to do to determine the absolute maximum and the absolute minimum f x12x3 48x2 36x 0 12x3 48x2 36x 0 12x3 48x2 36x 0 12xx2 4x 3 0 12xx 3x 1 critical points 0 l 3 absolute minimum cr1t1cal pomts absolute maximum endpoints 27 Calculus Refresher Comse Fall 2007 We conclude that fhas an absolute maximum of 37 and an absolute minimum of 727 on the interval 71 4 The graph of f con rms our conclusions see gure The Mean Value Theorem Iffis differentiable on a b then there exits a number 0 between a and b such that fc w ie there exits a point c f 6 Where the instantaneous a slope equals the average slope from a f 0 to b f b L Hospital s Rule Supposefand g are diITerentiable and g39x 0 near 0 except possibly at 0 Suppose further that limin has indeterminate form or 9 Then Ha g x 700 ME h ffx Ha goo Ha g x provided the limit on the right exists Note The above statement of L Hospital s Rule omits some details For exam 1 L Hospital s Rule also applies in a similar Way to limits for Whichx approaches 700 or 00 instead of x approaching a real number 0 Also L Hospital s Rule applies When the limit on the right does not exist so long as it is 00 Calculus Refresher Course Fall 2007 sin x Example 3 Evaluate lim xgt0 x We cannot evaluate the limit by direct substitution because the function is not continuous at 0 because it is not de ned at 0 Nonetheless the function may have a limit at 0 We cannot know for sure without further analysis Since both the numerator and the denominator approach 0 as x approaches 0 the limit has indeterminate form The notation does not indicate the limit does not exist it merely indicates the existence of sin x the limit cannot be determined with out further analysis To further analyze lim we xgt0 x apply L Hospital s Rule sin x cos x cos 0 11m 111117 7 xgt0 x xgt0 1 1 1 Example 4 Evaluate limeiz Hoe x Here we have a limit with indeterminate form 3 L Hospital s Rule applies lim 1imi 1imi oo Hoe x2 Hoe 2x Hoe 2 L Hospital s Rule is applied a second time x e because limi also has 1ndeterm1nate form g Hoe 2x 29 Calculus Refresher Course Fall 2007 Example 5 Evaluate lirrsifid H n x This is a trick problem The limit does not have indeterminate form the numerator approaches 0 but the denominator does not In fact the function is continuous at 5 and so the limit can be evaluated by direct substitution x 55 5 limi i0 H5 lnx ln5 If one were to erroneously apply L Hospital s Rule a different incorrect answer would be found hmfi3umlhmx5 xgt5 nx xgt5 xgt5 Hm L Hospital s Rule is erroneously applied eventually leading to an incorrect answer Take some time to respond to the following items 1 Find open intervals on which the graph of the given function f is increasingdecreasing also find open intervals on which the graph of f is concave upwarddownward also find all relative maximumsminimums and all in ection points Use this information to sketch the graph off fx1027x x3 30 Calculus Refresher Course Fall 2007 2 Find the absolute maximum and the absolute minimum of the given function on the given interval fx x JE 04 34 Evaluate the limit be careful to apply L Hospital s Rule only where it is applicable sin x 11m xgt7r x2 E 2 cos x Ho ex 12 4 x 11m 3 lenle 31 Calculus Refresher Course Fall 2007 17 VII The De nition of the De nite Integral I f x dx of a Function f Recall the de nition 17 f x d 3 lim 2 f xjAx 0f the de nite integral of f on the interval a b a Mac 11 The 2 portion of this notation signi es a sum consisting of several terms 71 indicates the 11 number of terms in the sum One can see the de nite integral of a function f is de ned in terms 0fthe limit 0fthe sum 2 fxAx This sum is called aRiemann sum The notation ZfxAx 11 11 is used to more ef ciently represent fxf Ax fx Ax fx Ax fx Ax The 7 terms in a Riemann sum correspond to n subintervals of equal width of a subdivision of ab see the gure Further xf x xg x denote sample points within the respective subintervals The interpretationmeaning V of a Riemann sum and subsequently of a r de nite integral depends on the r contextmeaning 0f the function f and 0f glib V g A if M I T the quantity Ax Consider the following I quotAquot K x 5 three examples of various possible contextsmeanings 0f the function f and the quantity Ax 53 x 1 39 in 51 Example 1 Suppose f x gives the linear density grams per meter of a straight wire along the xaxis from a to b and that Ax represents the length m of each of 71 sections of the wire In this context each term f xAx 0f the Riemman sum represents the approximate mass grams per meter times meters meters of the corresponding section of wire Therefore in this context the Riemann sum 2 f xAx approximates the total mass of 11 the wire Example 2 Suppose fx gives the force lb on an object that is moved along the x axis from a to b and that Ax represents the length ft of each of 71 sections of the path the obj ect follows In this context each term f xjAx 0f the Riemman sum represents the approximate work ftlb done across the corresponding section of the path Therefore in this context the Riemann sum 2 fxAx approximates the total work done in moving the object 11 from a to b 32 Calculus Refresher Course Fall 2007 Example 3 Suppose fr is merely regarded as theycoordinate paired with r ofa point on the graph offthat lies entirely above the raxis In this context each term xAr o the Riemman sum represents the area ofa rectangle which approximates the area of the corresponding section of the region bounded by the graph offand the raxis see figure Therefore in this context the Riemann sum ZfQUAX l O a x b 1 total area of the region bounded 39 by the graph offand the raxis from a to b 39 in each of L L 39 of some sort by chopping into pieces 39 39 39 for earh niere and 39 39 39 can the individual pieces There are many other contexts that we could consider Whatever the context generally the Riemann Sums better approximate the total as the the number n of sections increases Therefore the exact total is the value the Riemann sum approximations approach as n approaches so that is the exact total is given by rfrd liranrAr Accordingly iffis a density function for a wire then 11 Ha H Ibfrdr is the exact total mass ofthe wire iffis a force function then Ibfrdr is the exact total work Geometrically speaking r f rdr gives the exact total area of the region bounded by the graph ofa nonnegative function fand the raxis from ato b In the case where the graph offdips below theraxis rfrdr gives a net area VIII 39 39 39 lndP nitP lntPanls Thewnrd quot 39 39 39indicativeofL 39 An quotquot 39 quot fafunction fis a function whose derivative isf For example an antiderivative of Zr is rz since Ex Zr Notice rz 3 rz 7 Ir andr2 17 are also antiderivatives of Zr since each oftheir derivatives is also equal to Zr In fact Zr has infinitely many antiderivatives each of which can be wrinen as rz C for some constant C The family of all antiderivatives of rz can be expressed as rz C Another notation for the family of all antiderivatives of Zr is IZrdx Using this new notation wewriteIZrdxr2C Calculus Refresher Course Fall 2007 In general the inde nite integral off denoted by I f x dx is de ned to be the family of all antiderivatives of f in this situation f is called the integrand Because antidifferentiation is essentially the reverse of differentiation for every derivative ruleformula we have there is a corresponding indefinite integral ruleformula For example disin x cos x ltgt cos x dx sin x C The indefmite integral formula cos x dx sin x C x and scads of others are found on reference pages 48 in the back cover the text book The most basic frequently used formulas are found on reference page 4 see below Some of these are obvious reversals of basic derivative formulas To derive others such as F12 and F14 special techniques we have yet to review are needed However these formulas are easily veri ed by differentiating the rightside the derivative of the righthand side should equal the integrand of the lefthand side This rst group is the algebra of integrals This formula results from using integration by parts dx dx Icfoc Clfoc 100 IlnxdxxlnxixC Ifx goodx jfoc dx jgocmx Ifxgxdx Mxmxi gem TABLE OF INTEGRALS QASIC FORMS gt a 41 7 mm quotWm 39 lJudvm vzlu 1 llJ cscucotudu cscuC z39Jlundu quotH C 11 12 ltanudulnlseculC n l du l3cotudulnsinulC 3 i ln lul C u h k W MindululsecutanulC 4 e d e C N V ham T 15JcscudulnlcscuicolulC 5 I a du 107 C w 4 m 5 m quot9 w n a 14 u 16 quot quot C C fifwu v W Y I m sin a 1 l6JsinuducosuC d 1 i l 5 U 1 17tan1 Cl 7 Jcos u c111 sin u C i L a u a 3777 7 mg i l 1114 l u l8J7Sec l C 8 J r E M4113 e a3 a a 3939z quotA 19 d Zi1 quot a 6 a i u 20 u i a d l my m C u i a 20 u a 34 Calculus Refresher Course Fall 2007 Example 4 Verify formula F12 To verify formula F12 we must show 611 lnsec u C tan u Differentiating lnsec u u requires the Chain Rule dilnsecuC secutanutanu u sec u Example 5 Find the indefinite integral of the given function fx 6x5 3x4 l2x2 5 jfxdx 68 3x l2x2 5dx 68dxj3x4dx j12x2dxj5dx 6 5 3 6ij dx3jx4 dx 12Ix2 dxj5dx 613i 1215xc 6 5 3 J 3x5 v 6 3 x 4x 595 C At this step the constants of integration for the four integrals on the lefthand side are lumped Therefore I f x dx x6 4x3 5xC Fogethef as one cougar 0f 5 1ntegration on the r1ghthand s1de With a little practice every calculus student should be able to integrate any polynomial almost automatically without going through all the intermediary steps shown above The next example is similar with some of the steps omitted Example 6 Find the indefinite integral of the given function fx x7 8x3 l2x 8 4 Z 8 x dx x7 8x3 l2x dx L 8i12ic 1 2x 6x2 C If 8 4 2 8 Example7 Find the indefinite integral of the given function fx3e 6sinx l x Ifxdx 136 6sinx ljdx3e 6cosx ln x C x 35 Calculus Refresher Course Fall 2007 17 The concept underlying de nite integrals ie I f x dx represents total mass force area etc and the concept underlying inde nite integrals ie I f x dx represents a family of antiderivatives are two very different seemingly unrelated concepts However the similar notation and vocabulary are not coincidental The Fundamental Theorem of Calculus establishes a remarkable and very useful connection between these concepts IX The Fundamental Theorem 0f Calculus Time does not allow us to delve deeply into The Fundamental Theorem 0f Calculus FTC Suffice it to say that if a function f is continuous on ab then f f x dx exists and its value can be found by finding any antiderivative F of f and calculating F b F a That is jfx dx Fltb Flta where F is any antiderivative of f The remarkable thing about the FTC is that it indeed establishes a relationship between the two seemingly unrelated concepts of toLal mass force area etc and antiderivatives The useful aspect of the FTC is that it provides a simple way for evaluating definite integrals without having to face the seemingly impossible task of finding the limit of a Riemann sum Using the FTC to evaluate 17 f x dx is only simple provided one can find an antiderivative off Such a task is often not simple at all Reference pages 48 of the textbook provide formulas for finding antiderivatives of lots of functions However this list is very small considering the infinitely many functions it does not provide antiderivatives for Nonetheless the given formulas in conjunction with techniques reviewed in subsequent sections will allow us to accomplish quite a lot in terms of evaluating definite integrals via the FTC For now let s evaluate some relatively basic definite integrals Examplel Evaluate x2 dx 3 x3 3 x3 Now is an ant1der1vat1ve of x2 and so J1 x2 dx g 3 3 x where 1s a notation that 0 3 x reminds us to evaluate at 3 at 0 and to subtract Therefore 3 33 Ixzdx 0 30 3 3 il 9 3 3 3 Example 2 Evaluate Ildx x 1 36 Calculus Refresher Course Fall 2007 Ildxlnlx lne lnll 0l 1x 2 1 5 Example 3 Evaluate J cos xdx 7r 6 5 1 1 3957r397r 00sxdx s1nx s1n s1n 2 20 Take some time to respond to the following items 13 Find the inde nite integral I f x dx of the given function f l fxx5 3x2 6x4 2 fx 4sinx 7 3 fx3e cosx sec2 xi x 37 Calculus Refresher Course Fall 2007 46 Evaluate the given de nite integrals 4 fx3 6x5dx 5 Asecxtanxdx 0 6 IA de 1 x 38 Calculus Refresher Course Fall 2007 X Techniques of Integration Recall that evaluating de nite integrals is relatively easy via the FTC provided one can nd an antiderivative of the integrand In many cases nding antiderivatives is not as simple as looking up a formula in an integral table The techniques reviewed in this section are often used to nd antiderivatives A The Substitution Method Recall that for every derivative formula there is a corresponding integral formula The substitution method is a technique of integration that corresponds to the Chain Rule i gwl f39gxl39g39x G If39gxl g39x fgxlC Chain Rule Basis for the Substitution Method A clever use of notation shown below simpli es the technique and justi es the name Substitution Method sometimes called Ltsubstitution j f39gx g39xdx Mum LY Y u du The general idea is to reduce a complicated integral to one that is less complicated by substituting u for a function often inside another function in the integrand whose derivative is also a factor of the integrand Example 1 Use the substitution method to nd Jsin x2 2x dx Let u x2 Then du 2x dx After making these substitutions we have sinxz xdx Isinu du cosu C cos x2 C Therefore Jsin x2 2x dx cos x2 C One can check their work by differentiating the righthand side If all is done correctly the derivative of the righthand side equals the integrand function of the lefthand side Check it cos x2 C sin x2 2x sin x2 2x 39 Calculus Refresher Course Fall 2007 Example 2 Use the substitution method to nd ake x Let u In x 1 Then du idx x After making these substitutions we have I dx16dulu du2uC3lnxC x 3 3 Example 3 Use the substitution method to nd Ixzexzdx Let u x3 Then du 3x2 dx In this case we do not have all the elements of du in the original integrand function We can however tweak the integrand so that we have all the elements needed to perform the substitution Ixzexzdx lJ3xzexzabc lIeWu 1e C 1 C 3 3 3 3 77 Example 4 Evaluate J AsinS xcos xdx Notice this integral is a definite integral We can evaluate it if we can find an antiderivative of the integrand function We will use usubstitution to find the antiderivative When we make the substitution we will also substitute the bounds of integration with bounds consistent with the substitution 5 77139 whenx1s Let usinx 3 uisO henxisnuis Then du cosxdx 7 7J2 I7Asin3xcosxdxIo MS du 7 40 Calculus Refresher Course Fall 2007 J Example 5 Evaluate I 3 dx x J J dx29e 13 425 B Integration by Parts dc2je alu2e 2e3 2e2 Zeze l The technique referred to as Integration by Parts is a technique of integration that corresponds to the Product Rule d a fxgx f39xgx fxg39x cgt j f xgx fxg39x d F fxgx C cgt j f xgxd xr j fxg xd F fxgxC cgt j fxg39x dx g 1 gm f x dx dv u v v As with usubstitution clever notation simpli es the process of integration by parts Iu dv uv Ivdu See formula F1 on p 35 0fthis document 41 Calculus Refresher Course Fall 2007 Example 6 Evaluate Ixe dx Let u x and dv e dx Then duldx and ve After making the substitutions we have Ixe dx Iudv uv Ivdu xe Je dx xe ex C Therefore Ixe dx xe 6x C We can check our answer by differentiating the right hand side 1xe ex Cle xe ex xex dx Example 7 Evaluate lnxdx Let u lnx and dvdx l Then duldx and v3x x 3 After making the substitutions we have lnxdxlnx3x JEx ldx3lnxx Ejxdx 3 3 x 3 3 Elnxx E2xCElnxx ixC 3 3 3 3 9 42 Calculus Refresher Course Fall 2007 MM Evaluate 1x2 sin x dx Let u x2 and dv sin xdx Then du2xdx and v cosx After making the substitutions we have 1x2 sin xdx x2 sinx I cosx2xdx x2 sinxJ2xcos x dx To continue we apply integration by parts a second time Let u 2x and dv cos xdx Then du 2 dx and v sin x Now we have 1x2 sin xd JclcsinxI2xcosx d Jclcsinx2xsinx Jsinxd xzysinx 2xsinxcosxC C The Method of Partial Fractions The Method of Partial Fractions is useful for nding antiderivatives of rational functions recall a rational function is a ratio of polynomials Review for a moment the process for adding rational functions that involves finding a common denominator etc 1 2 l x 3L 2 x x 3 L 2x x 32x 3x 3 x x 3 xx 339x 3 x xx 339xx 3 xx 3 xx 3 Therefore li 3x 3 Then I 3x 3 dx can be found by evaluating x x 3 xx 3 xx 3 Ii dx which is doable Of course we had the advantage of the advanced x x 3x 3 xx 2 1 functions 7 amp 7 x x 3 knowledge that can be decoupled into the sum of two easier to integrate The Method of Partial Fractions is a method for decoupling complicated rational functions into easier to integrate functions which are not known in advance The method will always work so long as the degree of the numerator of the rational function is lower than the degree of the denominator if it is not the situation can be rectified by long division Perhaps the technique is best described by example 43 Calculus Refresher Course Fall 2007 5 Example 9 Evaluate J x2 9 dx x 5 Before we worry about the integral we focus on decoupling the fraction x2 9 into x partial fractions If the degree of the numerator is less than the degree of the denominator the first step is to factor the denominator x S x S x2 9 x 3x3 The partial fractions must have denominators whose common denominator is x 3x 3 We therefore conclude the partial fractions must be 7 7 x 3 x 3 If the sum of these fractions is to have a numerator of lower degree than the degree of the denominator then the partial fractions themselves must both have numerators of lower degree than their respective denominators In this case we conclude both numerators must be constants Therefore x 5 ii for some unknown constants A and B To find these x 3x3 x 3 x3 unknown constants we first multiply both sides of the equation by the common denominator with the goal of clearing all denominators L5ii x 3x3 x 3 x3 x 3x3iix 3x3 x 3 x3 x 5Ax3Bx 3 x x 3x 3 At this stage suitable substitutes for x allow one to solve forA and B x333 5A33B3 33 26A3 x 3 3 5A 33B 3 3 8 6B3 x 5 Ax3Bx 3 1 x5 x 3x3 x 3 i Therefore i x 3 44 Calculus Refresher Course Fall 2007 Now we return to the original inde nite integral problem Ix ddxlj xiIdeiln x 3liln x3lC x 3x3 3 x 3 3 x3 3 3 The integrals were evaluated Via usubstitution with details omitted 10 Example 10 Evaluate dx J x3 3x2 40x 10 10 10 AL B C J x3 3x2 40xxx2 3x 40 xx 8x5 x 39 x 8 39 x5 10 A B C 4747 xx 8x5 x x 8 x5 10 A B C Wxx 8x5 Mmxx 8x5 10 Ax 8x5Bxx5Cxx 8 x0310 40A3A 1 I 10 Ax 8x5Bxx5Cxx 8 x 8 10 104B 2 B 2 VI 3 3 x 51065CC 10 11 5 1 2 1 Ixx 8x5 27b lx 8 El x ilnlxlilnlx 8lllnlx5lC 4 52 13 45 Calculus Refresher Course Fall 2007 Take some time to respond to the following items 15 Use the substitution method to evaluate the given de nite and inde nite integrals l sinxcosxdx H 2 Iede 3 Jde J 4 J1 gxcosxzdx 46 Calculus Refresher Course Fall 2007 68 Use integration by parts to evaluate the given de nite and inde nite integrals 6 Ixcosxdx 7 Ixze dx 8 fxlnxdx 47 Calculus Refresher Course Fall 2007 910 Use partial fractions to evaluate the given de nite and inde nite integrals x10 9 761x Jx3 25x a 10 10 L mdx 48 Caleulus Refresher Course Fall 2007 Consrder the problem of ndmg the area of the unbounded shaded reglon see gure Srnee the reglon ls unbounded some people39s rnturuonleadthern to predretthe arears lnflnlte Sueh ls not the ease The area lnflnlte or not ls approxlmated by the 1 x 5 gt1 The aeeuraey ofsuch an o de nrte rntegral 1 x approxlmatlon rnereases as d rnereases The enaet arears the nurnberthe approx rnauons approaeh as d approaehes oo Thus the areaA ofthe shadedreglon ls glven by o 1 e Allm17dxllm7 M l X M X The preyrous dlscusslon rnotryates the followlng de nruon The Impropermlegral offfrom a to do denotedby rm ls deflnedln terrns ofthe hrnrt of a deflnlte rntegral as follows a o L rdrlrrn LXW 10 of the textbook Examgle 1 Evaluate I By de nruon the value ofan rrnproperrntegral ls found by eyaluaung the lrrnrt of a de nrte rntegral x 1 0 2x 1 at 1 dxllm7 idxrllm 7 X272 p121 X 2 2M7 J longs7 jenny amemm du 7 2 dz hrn J M Calculus Refresher Course Fall 2007 Example 2 Evaluate I e39x dx no l7 Z7 717 J equotdxllmJ equotdx limJ equotdx 11mJ e du 0 l7gtoo 0 l7gtoo 0 l7gtoo 0 717 b 0 l 0 12e e 12g1 1 lim 6 Take some time to respond to the following items 12 Evaluate the improper integral 1 I xe39xzdx Calculus Refresher Course Fall 2007 XII Outline of Calculus II Formulas CHAPTER 6 De nition of a De nite Integral 17 f xd a limZZ1 f xjAx ngtoo 17 Area I f xdx is the area above the xaXis subtract the area below the xaXis f7 f x gx dx is the below the graph of f and above the graph of g Where x gt gx Volume V 17 Axdx area of a crosssection methodithe disc and washer methods are special cases of this method V I IL39fx2 dx disc method V 7139fx2 g x2dx washer method V 2mg xdx shell method Arc Length L Wake rectangular coordinates L dt parametric equations in two dimensions 1 Average Value fm b a I fxdx Applications to Physics and Engineering M W Fxdx work done in moving an object from a to b IV ngId work done in pumping a slice ofliquid Hydrostatic Force E pgATdI force of a slice of liquid on a vertical surface Moments Mass and the Center of Gravity M x 17 17 2 17 7 My 7 Mypjaxfxdx Mxpja7fxdx mplafxdx F7 y m m Calculus Refresher Course Fall 2007 CHAPTER 7 Euler s Method xquot x 1 h yquot y 1 hFxn71yn71 Exponential GrowthDecay Pt Poe CHAPTER 8 Series in General S lim squot where squot a1 a2 an sum ofthe series 2161 Geometric Series Zoo 1611quot 1L l lt r lt1 quot r Remainder Estimates onl f xdx S Rn S J 0 f xdx for series that converge by the integral test Rn S b 1 for series that converge by the alternating series test Also Taylor s inequality given below no H Taylor Series of the function f centered at a Zfifmw aquot n0 7 M quot1 g 77 n1 Tayor s Inequality Rn x S W x a where M 1s a bound on f n t u M Taylor Polynomials Tn x fa flea x a fzaa x a2 f7ax aquot n Calculus Refresher Course Fall 2007 CHAPTER 9 AND APPENDIX G Distance d x2 JC12 y2 y12 Zz 21y Sphere x h2 y k2 z l2 r2 Vectors Magnitude a la a22 a Unitize unit vector in the direction of a a Dot Product De nition ab a b cos 639 For Computation ab alb1 azb2 a3b3 ab Application W F D work cos 639 W a b b b b Projections compab L prOJab L 1 aiza La la W W lal 3393 Cross Product 1 J k De nition a X b a b sin 6n For Computation a X b a1 a2 a3 Application T rF torque A 11 X V area of a parallelogram V la b X c volume ofa parallelopiped x x0 at Lines and planes y y0 bt parametric equations for a line 2 20 ct ax by cz ax0 by0 cz0 equation for a plane D lax1by1czl ax0 by0 czol Ia2 b2 c2 distance from a point to a plane 53 Calculus Refresher Course Fall 2007 Polar Coordinates Conversion Formulas x r cos 6 y r sin 6 polar t0 rectangular r2 x2 yz tan 6 Z rectangular to polar x dy 2 Slope m Area A I lf62 d6 Arc Length L V r2 i 616 dx 1 2 391 d6 d6 Cylindrical Coordinates Conversion Formulas x rcos 6 y rsin 6 z z cylindrical t0 rectangular y r2x2y2tan6 x z z rectangular t0 cylindrical Spherical Coordinates Conversion Formulas x p cos 6 sin y p sin 6 sin 2 p cos spherical t0 rectangular p2 x2 y2 x2 tan 6 Z cos i rectangular t0 spherical x p Calculus Refresher Course Fall 2007 CHAPTER 10 Calculus of Vector Valued Functions Theorem tie back to real Intuitive Topic De nition valued functions concept limit 9fa For all 8 gt 0 there exists lim r0 AS I functlon 5 gt 0 such that Ha ptproaches a lim tlim tlimhtgt 121111 0 L 0 lt t a lt 5 3 1 ILl lt 8 ltHa Ha g Ha approaches L 1 is continuous at I a if a functlon that 1 r is de ned at t a 1 1s cont1nuous at I 61 1f the No jumps is component functions f g and holes breaks 11m1 t cont1nuous at a 2 Ha ex1sts h are all cont1nuous at I 61 etc where pomt 3 12121 1a t a derivative of a Tangent vector function 1quota limW 1quota ltf a g39a h39agt vector at a point HO h velocity 17 b n J raw Related to the defin1te 1ntegral J 1 tdt limZ 1 t1 At b b b position H In ftdt jg gown a hltrgtdrgt andor d velocity ofa inde nite J raw denotes the family Ira t pa ide in Integral of all antiderivatives off ltj ftdt j gtdt j htdtgt who Smoothness A curve described by a vector function 1 is smooth at t a if 1quott 0 Are Length 117 1quott dt r39I Key Umt Vectors Tt Nt r39I dT T39tl 1quott gtlt 1 quott Curvature K 39 3 ds 1 0 r39t T39a To Bt Tt gtlt Nt fquotx 1 f x2 13 Calculus Refresher Course Fall 2007 Motion m r0 Velocity Vt 1quott Speed vt vt Acceleration at V39t rquott a aTT aNN where air and cm are given below Va r39trquott aT v 7 39 tangential component of acceleration v 1 0 2 r39tgtltrquott aN KV W normal component of accelerat10n r

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All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.