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# Calculus 2 MATH 156

WVU

GPA 3.8

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This 12 page Class Notes was uploaded by Rae Kutch on Saturday September 12, 2015. The Class Notes belongs to MATH 156 at West Virginia University taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/202663/math-156-west-virginia-university in Mathematics (M) at West Virginia University.

## Reviews for Calculus 2

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Date Created: 09/12/15

Review topics for exam 1 1 Differentiate an integral using the fundamental theorem of calculus Here you can just remember the formula i W a dx mmdr f dx fox dx Comment Yes you would get this answer if you imagined that Fx I xdx was an antiderivative so that you could calculate using the chain rule 1 i da dx lfrgtdr dx Fm For F m F am f dx fox dx the same answer The point of the fundamental theorem of calculus is really that an antiderivative exists and is constructed via the definite integral namely Fx a ndt satisfies F x fx 2 Average value of a function 3 Basic integration formulas This is just your basics such as Mdu 1 M 11 1 n 1 Isinau du cos au Icos au du sinau 18 st e j du lnlul 14am tan lu 1 u2 Isecu du mlsecu tanul Itanudu 1nlcosul lnlsecul but you can easily derive this one 4 Simple integration problems substitutions may often result in these types of integrals 2 2x12 jx2x1dxj1f2xdx12x1 dxIC 2dx 1 5 Substitutions you can substitue anything Simple forms sinx L 1 200 dx Is a I u du problem Ixe zxzdx is a Ie du problem dx is a Isinu du problem I sinkY J 1117 dx is a judu problem Slightly more involved x2 Ilzx dxuseu1 2x 1 dx use u 1 x2 but what if the numerator were x2 instead of x3 x Ix3e zxzdx 6 Integration by parts judv uv Ivdu also the definite integral version Standard examples for which integration by parts is effective kasinx dx kacosx dx kae dx kalnx dx Isin lx dx could substitute u sin lx or integrate by parts directly Ix41 x5dx Use integration by parts to show thatsze x2dx xe 2 8quotde The following integral shows up in partial fractions dx Using integration by parts k 1 ll xz x2 l dx law dx i dx j 2 x k dx 1x2 1 1 1 x2k 1 HZ 1 1 x 1 Ilt1x2gtk 1dx 2010 mm 21 kI1x2k 1dx 2k 3 I1x2k dx 2k1 1x2k 1 2k2 I 1x2k 1dx k gives us a direct reduction formula rather than resort to trig substitutions 7 Trig integrals Basic identities sinzx coszx 1 1 tanzx seczx Odd positive powers of sine or cosine even powers of secant odd powers of tangent Even powers of sine and cosine odd powers of secant reduction formulas 8 Trig substitutions Basic quadratic forms and associated substitutions substituting back Completing the square 9 Partial fractions I p00 dx qx Divide first if degp lt degq does not hold Factor denominator into real linear factors and irreducible quadratic factors quadratic factors with complex roots The form of the partial fraction decomposition Determining the coefficients can be done by substitution or by equating powers after multiplying both sides by the common denominator qx How each term can be integrated 10 Integrals that result in rational functions upon substitution eg 5 dxuc 11 Using integral tables simple substitutions like u ax Completing the square Using reduction formulas Math 156 Topics for Exam 4 00 Power series 2 cnx aquot power series about x a or centered about x a quot0 1 1 x w A basic power series representation 1 x x2 2xquot lxl lt 1 quot0 Determining radius of convergence and interval of convergence of a power series Finding the derivative and integral of a function defined by a power series Manipulating power series eg if 8 1x x22 x33 00 lt x lt 00 what are power series in powers Ofx for e 8quot 8 2 Iex2dx exx 1 I etc 1 dx what is the power series for ex aboutx 1 Taylor series If a function has a power series about x a that is valid forx in some open interval aboutx a then fx 2 Wowquot n0 and the series on the right is called the Taylor series offabout x 61 Developing Taylor series for various functions eg sinx cosx ex J ll x 1nxln1x 1 n k Taylor polynomial degree n Offx about x a denoted Tnx Tnx Z x ak k0 The remainder Rnx Estimating the remainder via Taylor s remainder theorem f quot1gtz RM n1 In an alternating series with decreasing terms the remainder can be estimated as the first neglected term in the Taylor series namelyfx is between Tnx and mx WWquot x aquot1 so the error fx Tm is between 0 and MWquot x art1 which n 1 n 1 is the quotnext termquot in the Taylor series offx So for instance if lxl we have x aquot1 where z is somewhere between a and x 2 4 cosx 2 1 X With error no larger than Parametric curves Sketching a parametric curve associating points on the curve with values of t Finding derivativeslope of tangent for a parametric curve Locating relative maxima Getting an equation in x and y by eliminating t can only do that sometimes Parametric equations of circles ellipses 2 2 Arc length of a parametric curve ds dt Area between a curve and the xaxis Iydx IQytdt where xaya is at the left and x y is at the right of the curve Formula sheet A function F x is an antiderivative of x if F x fx The inde nite integral of f is f f xdx F x C where F is an antiderivative off Fundamental Theorem of Calculus Assume f is continuous on 61 b Then Part 1 gx f f 0dr is also continuous on ab and gx is an antiderivative of f 7 Part 2 HF is an antiderivative off then ffxdx Fb Fa 7 Average value of a function fave 1 f f xdx b a a Selected properties of integrals pq and a b are constants fpfoc mm pffxdx quoodx b b b fpfoc mm pffxdx Mom 4 b a ffxdx 0 ffxdx ffxdx a a 7 Selected integrals n1 ffx gxdx ffxdx fgxdx fcfxdxcffxdx fxquotdx x C nl n1 fldxlnixiC fe dxequotC fa dx a C x lna fsinxdx cosxC fcosxdxsinxC fseczxdxtanxC fcsczxdx cotxC fsecxtanxdxsecxC fcscxcotxdx cscxC f 21 dxtan 1xC f 21 dxllnx71C f 1 dxsin 1xC x 1 x 1 2 x1 1x2 Substitution we denote ugx 7 5M ffgxg xdxffudu ffgxg xdx ffudu a ga Integration by parts 7 ffxg xdx fxgx f f xgxdx ffxg xdx fxgx b ff xgxdx Formula sheet Trigonometry Fundamental identities csc0 sec0 tan0 sme cot0c se sm 0 cos 0 cos 0 sm 0 cot0 10 sin20cos201 1tan20sec20 1cot20csc20 tan sin 0 sin 0 cos 0 cos0 tan 0 tan 0 cot 0 cot 0 sin 0 cos0 cos 0 sine tan 0 cot 0 cot 0 tan0 2 2 2 2 Addition and subtraction formulas tan x tan y s1nx y s1nxcosy cosxsmy cosx y cosxcosy s1nxs1ny tanx y 1 tan x tan y tan x tan y s1nx y s1nxcosy cosxsmy cosx y cosx cosy s1nxs1ny tanx y 1 tan x tan y Half and doubleangle formulas 2 2 2 2 2tanx s1n2x 2s1nxcosx cos2x cos x s1n x 2cos x 1 1 2s1n x tan2x 72 1 tan x 1 1 sm2 x 51 cos2x cos2 x E1 cos2x Derivatives of direct and inverse trigonometric functions 6sinx cosx 6cosx sinx aanx sec2 x d d d 2 icscx cscxcot x isecx secxtanx icot x csc x dx dx dx d 1 1 d 1 1 d 1 1 ism x icos x itan x7 dx dx 1x2 dx 1c2 1csc391x 1 1sec391x 1 1cot391x 2 dx x 1x2 dx x 1x2 dx 1x Areas and volumes Area between two curves A ffx gxdx Volume formulas xmax xmax Rmax ngm fAxdx VRM fnR2x r2xdx VW f2nfxxdx Rmm quotquot2 Path length L f11 f x2dx xmm Math 156 Review topics for Exam 2 Improper integrals What makes an improper integral improper The definition of the value of an improper integral in terms of limits evaluating improper integrals this way Using L Hopital s rule to evaluate resulting limits xe xdx Iilnx dx Special improper integrals you should know and be able to show I L dx converges forp lt 1 diverges othenNise for example 1 1 1 1w Hm lox de 0pr dx converges forp gt 1 diverges othenNise for example 1 1 1 I 7 I f 117 I 8quot dx 1 converges lnx dx 1 converges xe x dx Comparison for positive integrands Roughly speaking if an integral converges any smaller integral converges if an integral diverges any larger integral diverges To analyze determine the dominant term in a sum when you approach the discontinuity or as you go to 00 That will usually tell you whether or not the integral converges co 1 11 ex 00 1 11 0 Io xl e dx oniff dx II xx2 dx Jolt1 dx Ioe dx Area between curves Pay careful attention to specification of the region in question You may need to find points of intersection yourself Depending on shape of the region it may be more convenient to use vertical strips and integrate with respect to x or to use horizontal strips integrating with respect to y Volumes By method of slicescross sections V IAxdx Typical problem We specify the base and describe crosssections perpendicular to one of the axes You figure out the dimensions of a general cross section and then the area Obviously you might find yourself integrating with respect to y as well as with respect to x Volumes of revolution Determine the region to be rotated The shape of the region ie the bounding curves and the functional form of those curves is y in terms of x can we solve forx in terms of y will usually suggest whether you want vertical strips integration with respect to x or horizontal strips integration with respect to y Then the axis of rotation will determine which of the following approaches applies Diskswashers When you rotate strips perpendicular to the axis of rotation you get washers or disks if one end of the strip is on the axis of rotation V 1703 riznyzx rotation axis of form y a such as x axis y 0 The integrand represents the area of a washer a disk with a hole removed times the thickness dx Cylindrical shells When you rotate strips parallel to the axis of rotation you get cylindrical shells V 2mde rotation axis of form x a such as y axis x 0 Above r is the distance of the strip from the axis 0 represents the length of the strip 0 ytop J bot when rotating about axis x c lntegrand represents area W of rectangular strip times circumference 2m of circle when strip is rotated In applying these formulas to specific problems we often make use of notation such as xright xle ytop ybot Ex Rotate region in first quadrant bounded byy x3 x y 0x 1 about the a x axis Here integrating with respect to x and using washers is very easy b y axis Integrating with respect to y and using washers is hard to do because we can t solve for x in terms of y lnstead integrate with respect to x and use cylindrical shells In this example the functional form of the curves is driving the integration method Region in first quadrant bounded byy x2 y 2x2 4y 0 This was on the quiz Here the geometry of the region suggests that using horizontal strips is easiest and you can do it because you can solve for x in terms ofy easily Region bounded byy x2 y 2x2 4x 2 0 Here the geometry is more suited for vertical strips Note that we sometimes want to rotate about a line like x 2y 3 etc and then you have to carefully identify the various lengths and radii in the formulas Arc length 3 I 1 2 dx Sometimes we write ds dx2 aiy2 and then this X1 can be converted into eitherthe integral at the left when integrating with respect to x or the integrals 2 l1 2 dy if integrating with respect to y 1 Work A force Fx that depends on position x exerted over some interval a g x g 13 produces work W IFxdx Examples include stretching a spring compressing a piston Work problems also appear eg when building a wall pulling a hanging chain up pumping out a liquid In these cases the material is to be redistributed in space and each little bit of material requires a little bit of work to move it where it is supposed to go Pressure and total force P pgh is the pressure at depth h of a liquid of density p where g is the acceleration of gravity in the appropriate units Force is pressure times area sometimes write w pg as the quotweight densityquot the number of pounds per unit volume when using English units feet pounds Becuase the fluid is typically distributed across a range of depths each piece of the fluid is exerting different pressure and the total force must be calculated using an integral generally using horizontal strips of area across which the pressure is constant Center of masscentroid Centroid M A area IxO top yboz x ytop ybot Each x coordinate is quotweightedquot with all of the area located between x and x dx and then you divide by the total weight 3c i dem i deA i IxQmp yb0tdx if using vertical strips 7 i ydm i yd1 i I W02 yb0tdx if using vertical strips in the integrand 7 represents the centroid of the vertical strip located at W Use analogous integrals when using horizontal strips You can sometimes make use of symmetry as the centroid must lie on any line of symmetry of the region As an example for a semicircle above the x axis of radius a equation x2 y2 612 you could use horizontal strips 2 7ra A 2 3 L 22 Ll 2232 2a3i y AIOyQJa ydy A 3a y 0 WZz Ma or vertical strips Lal La az xz L2a3 2a332a33i y A lede A L 2 dX A 2 7ra2237ra Try for the region beIOWy 4 3x2 and above y 2 x2 Math 156 Review topics for exam 3 Sequences sequence notation an1 an fn n 12 or simply a1a2 developing a formula for a given sequence even numbers odd numbers exponentials powers limits of sequences techniques such as factoring out the largest term in a sum L Hopital s rule combining sequences limit laws Comparison of basic function classes who goes faster to infinity Series as Notation 2 an a1 a2 a3 quot1 n The sequence sn 0 of partial sums squot 2a i0 w The sum of an infinite series what it means Liggsn S and we write Z 61 S i0 Convergencedivergence absolute convergence Any increasing sequence either has limit infinity or converges Hence A bounded increasing sequence must converge Similarly for decreasing sequence eg any decreasing sequence of positive terms must converge Special series Geometric series finding the sum of a geometric series as 2 1 p SCI ICS 7 quot0 Analysis of series for convergence Convergence is independent of any number of terms at the beginning of the series So any comparison that is quoteventuallyquot true ie is true for n large enough can be used If series 2 an converges must have an gt 0 quot0 Basic comparison for positive series Combining and breaking up convergent series Detailed tests lntegral test an fn fx positive and decreasing 2 an converges if and only if IT xdx converges quot1 Limit comparison test especially useful for algebraic terms Compare with a simpler series whose terms are the same quotsizequot Ratio test I 6221 I gt L then ifL lt 1 series converges absolutely L gt 1 means series diverges especially good for series that involve factorials Root test lanllm gt L then ifL lt 1 series converges absolutely L gt 1 means series diverges good for series that involve taking an expression to the 11 power Alternating series Such series can sometimes be analyzed by looking at absolute value of the terms Alternating series test hypotheses Convergence versus absolute convergence

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