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# MATH 1226 Exam II Study Guide / Over View of Integrals and Derivatives MATH 1226

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This 5 page Study Guide was uploaded by austinc1 on Saturday June 11, 2016. The Study Guide belongs to MATH 1226 at Virginia Polytechnic Institute and State University taught by Sohei, Yasuda in Summer 2016. Since its upload, it has received 12 views. For similar materials see 1226 Calculus of a Single Variable in Math at Virginia Polytechnic Institute and State University.

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Date Created: 06/11/16

Calculus Cheat Sheet Integrals Definitions Definite Integral: Suppose f x( ) continuous Anti-Derivative : An anti-derivative of f ( ) on [ ] . Divide a,[ ]nto n subintervals of is a function, F( ), such that F ( ) f x ( ) * width Dx and choose x frim each interval. Indefinite Integral :òf x( ) = F x ( ) b ¥ Then ò f ( )x = lim å f ( ix . where F x( ) an anti-derivative of f x ( ) a nﬁ¥ i= Fundamental Theorem of Calculus Variants of Part I : Part I : If f( )is continuous on a[ ]then u( ) x d f ( )t = u x( )Øu x ( ) g ( ) òa f( )dt is also continuous on [ ] dxò a º ß d x d b and g x( ) ò f( )dt = f ( ) dxò ( )f ( )t = -v ( ) Øvºx( ) ß dx a Part II : f ( )s continuous on [ ] , F x ( ) d u( f ( )t =u x( ) [u(x)]-v ( ) [v(x] dxò ( ) an anti-derivative of ( ) (i.e. F( )= ò f ( )x) b then òa f( )dx = F b( )F a ( ) Properties ò f ( ) g x( ) = ò f ( )x– g ò d( ) ò cf( )dx = c ò x( ), c is a constant b b b b b ò f( ) ( )x dx = ò f( )dx– ò g ( )x ò cf ( )x = c ò f ( )x, c is a constant a a a a a a f( )dx = 0 bcdx = c(b-a ) ò a òa b a b b ò af ( )x = - òb f( )dx òa f( )dx £ òa f ( )x b c b òa f ( )x = òa f( )dx+ òc f ( )x for any value of c. b b If f ( ) g x( ) a £ x £ bthen ò f ( )x ‡ ò g( )dx a a If f( )‡ 0 on a £ x £ b then bf ( )x ‡ 0 òa b If m £ f ( ) M on a £ x £ bthen m b(a £ ) òa f( )dx £ M b(a ) Common Integrals ò k dx = k x+c òcosudu = sinu+c òtanudu = ln secu +c n 1 n+1 ò x dx = n+1x +c,n „ -1 òsinudu = - cosu+c òsecudu = ln secu +tanu +c -1 1 2 1 1 -1 u ò x dx = ò xdx = ln x +c òsec udu = tanu +c ò a2+u2du = atan (a) +c 1 dx = ln ax+b +c secutanudu =secu+c 1 -1 u ò ax +b a ò ò a2-u2du = sin (a) +c lnudu = uln ( )u+c csc uot udu = - csc u +c ò ò e du = e +c òcsc udu = -cotu+c ò Visitttp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. u Substitution : The substitution u = g x w( ) convert b f g ( ) ( )x = g( )f( )du using ò a ( ) òg( ) du = g x( ). For indefinite integrals drop the limits of integration. 2 2 8 Ex. ò 5x cos x( ) ò 5x cos x( ) = ò 3cos ( )u 1 1 1 u = x 3 Þ du = x 3x Þ x dx = du 1 5 8 5 3 = 3in u( ) 1 = 3(sin ( )sin 1 ( )) x =1 Þ u =1 =1 :: x = 2 Þ u = 2 =8 3 b b b Integration by Parts : udvò= uv- vdu anò òa udv = uv a- òa vdu. Choose u and dv from integral and compute du by differentiating u and compute v using v = dv. ò Ex. xe-xdx Ex. 5ln xdx ò ò3 u = x dv = e -x Þ du = dx v = -e - x 1 -x -x -x -x -x u = ln x dv = dx Þ du = dx v =xx ò xe dx = -xe + e dx ò -xe -e +c 5 5 5 5 ò3 ln xdx = xln x 3 ò3 dx = (ln x ( ) ) 3 = 5ln ( )3ln 3 -( ) Products and (some) Quotients of Trig Functions n m n m For sòn xcos xdx we have the following : For òan x sec xdx we have the following : 1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and cosines using sin x =1-cos x , then use convert the rest to secants using the substitution u = cosx . tan x = sec x-1, then use the substitution 2. m odd. Strip 1 cosine out and convert rest u = secx . to sines using cos x =1-sin x , then use 2. m even. Strip 2 secants out and convert rest 2 2 the substitution u = sin x . to tangents using sec x =1+tan x, then 3. n and m both odd. Use either 1. or 2. use the substitution u = tan x. 4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2. and/or half angle formulas to reduce the 4. n even and m odd. Each integral will be integral into a form that can be integrated. dealt with differently. Trig Formulas : sin ( )= 2sin x ( ) x , ( ) 2 ( )= 1 1+cos 2( ) sin 2 ( )= 1 1-cos 2x( ) 2( ) 2( ) 3 5 sin x Ex. tòn xsec xdx Ex. ò 3 dx 3 5 2 4 cos x ò tan xsec xdx = tanòxsec xtan xsecxdx sinx dx = sin xsin dx = (sinx)2sin dx ò cosx ò cosx ò cosx = (sec x-1 se) xtan xsecxdx (1 cosx)2sinx ò = ò 3 dx (u = cosx ) 2 4 cos x = ò(u -1 u)du (u = secx ) = - (1u2)2du = - 1 2u2+u4 du 1 7 1 5 ò u3 ò u3 = s7c x- sec 5+c 1 2 1 2 = 2ec x+2ln cosx - cos x+2 Visitttp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions. a -b x2 2 Þ x = sinq b x -a 2 Þ x = secq a +b x2 2 Þ x = tanq b b b cos q =1-sin q 2 tan q = sec q -1 sec q =1+tan q 2 16 16 2 12 Ex. ò x2 4 9x2dx ı 4 2 ( 3cos q )dq = ò sin qd q 2 2 9sin ( 2c)sq x = 3sinq Þ dx = 3cos q dq 2 = ò2csc dq = -12cotq +c 4-9x 2 = 4-4sin q = 4cos q = 2 cosq Use Right Triangle Trig to go back to x’s. From Recall x = x . Because we have an indefinite substitution we have sinq = 3x so, 2 integral we’ll assume positive and drop absolute value bars. If we had a definite integral we’d need to compute q ’s and remove absolute value bars based on that and, ì x if x ‡ 0 4-9x 2 x = í From this we see that cot q = 3x . So, î -x if x < 0 2 2 16 dx = - 4 4-9x +c In this case we have 4-9x = 2cosq . ò x2 4-9x2 x Partial Fractions : If integrating P( )dx where the degree of P x ( )smaller than the degree of ò Q( ) Q x( )Factor denominator as completely as possible and find the partial fraction decomposition of the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the denominator we get term(s) in the decomposition according to the following table. Factor in Q x( ) Term in P.F.D Factor in Q ( ) Term in P.F.D A k A1 A 2 Ak ax+b (ax+b ) ax b + 2 +L+ k ax+b (ax+b ) (ax+b ) Ax+ B A x+ B 2 Ax+ B 2 k 1 1 +L+ k k ax +bx+c 2 (ax +bx+c ) ax +bx+c ax 2+ bx + k ax +bx+c ( ) 2 2 Ex. 7x +12x dx 7x +23x = A + B2+C = A(x +4+ B2 C)x-1) ò x-1)x +4) x-1 x +4) x-1 x +4 x-1 x +4) 7x2+13x 4 3x+16 Set numerators equal and collect like terms. ò (x-)(x+4) dx = ò x1 + x2+4 dx 2 2 7 x + 13 x = (+ B x ) C - ( x+ A-) 4 = 4 + 2x + 26 dx ò x-1 x +4 x +4 Set coefficients equal to get a system and solve = 4ln x - + 3 ln x2 + 4 +8tan -1 x to get constants. 2 ( ) ( 2 A+ B = 7 C - B =13 4A-C = 0 Here is partial fraction form and recombined. A= 4 B = 3 C =16 An alternate method that sometimes works to find constants. Start with setting numerators equal in 2 2 previous example : 7x +13x = A x +4(+ Bx+C ) ( )( x-1 .)Chose nice values of x and plug in. For example if x =1 we get 20 = 5A which gives A= 4. This won’t always work easily. Visitttp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Applications of Integrals b Net Area : f( )dx represents the net area between f x ( ) the ò a x-axis with area above x-axis positive and area below x-axis negative. Area Between Curves : The general formulas for the two main cases for each are, b d y = f( )Þ A = ºpper functß-nºower functßdx & x = f y( )A = ºright funß-iºleft funcßdyn òa òc If the curves intersect then the area of each portion must be found individually. Here are some sketches of a couple possible situations and formulas for a couple of possible cases. b A= df ( ) ( ) dy c b A= òa f ( ) ( ) dx òc A= òa f ( ) ( ) dx+ òc g( ) ( )x dx Volumes of Revolution : The two main formulas are V = A x dxòan( ) = A y dy . Hòre ( ) some general information about each method of computing and some examples. Rings Cylinders 2 2 A=p (outer rad-usinner rad)us A= 2p radiuswidth / height Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl. to x/y of outer cyl. Horz. Axis use f x( ) Vert. Axis use f ( ) Horz. Axis use f y( ) Vert. Axis use f ( ) g x , A x and dx. g y , A y and dy. g y , A y and dy. g x , A x and dx. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Ex. Axis : y = a > 0 Ex. Axis : y = a £ 0 Ex. Axis : y = a > 0 Ex. Axis : y = a £ 0 outer radius :a- f ( ) outer radius: a + g ( ) radius :a- y radius : a + y width : f y( )g y ( ) inner radius : a- g ( ) inner radius: a + f ( ) width : f ( ) g y( ) These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the y = a £ 0 case with a = 0 . For vertical axis of rotation x = a > 0and x = a £ 0 ) interchange x and y to get appropriate formulas. Visitttp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Average Function Value : The average value Work : If a force ofF x( )ves an object b b of f ( )n a £ x £ bis favg= 1 f( )dx ina £ x £ b, the work done is W = òa F ( )x b-a òa Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, b b b L = ò ads SA= òa 2p yds (rotate about x-axis) SA= òa 2pxds (rotate about y-axis) where ds is dependent upon the form of the function being worked with as follows. dy 2 dx 2 dy 2 ds = 1+ (dx dx ify = f ( ) ,a £ x £ b ds = (dt +( dt dt if x = f ( ) y = g ( ) a £ t £ b dx 2 ds = r + ( )r 2 dq if r = f ( ) a £q £ b ds = 1+ (dy dy ifx = f ( ) , a £ y £ b dq With surface area you may have to substitute in for the x or y depending on your choice of ds to match the differential in the ds. With parametric and polar you will always need to substitute. Improper Integral An improper integral is an integral with one or more infinite limits and/or discontinuous integrands. Integral is called convergent if the limit exists and has a finite value and divergent if the limit doesn’t exist or has infinite value. This is typically a Calc II topic. Infinite Limit ¥ t b b 1. ò f( )dx = lim ò f ( )x 2. ò f( )dx = lim ò f ( )x a tﬁ¥ a -¥ tﬁ-¥ t 3. ¥ f x dx = c f x dx + ¥ f x dx provided BOTH integrals are convergent. ò-¥ ( ) ò-¥ ( ) òc ( ) Discontinuous Integrand b b b t 1. Discont. at a: ò f( )dx = lim ò f( )dx 2. Discont. at b :ò f ( )x = lim ò f( )dx a tﬁa+ t a tﬁb- a b c b 3. Discontinuity at a < c < b: òa f( )dx = òa f( )dx+ òc f( )dx provided both are convergent. Comparison Test for Improper Integrals : If f x ‡ ( ) ‡ 0o( ),¥ then[ ) ¥ ¥ ¥ ¥ 1. If ò f ( )x conv. then ò g( )dx conv. 2. If ò g( )dx divg. then ò f ( )x divg. a a a a ¥ 1 Useful fact : Ifa > 0 then òa xp dx converges if p >1 and diverges for p £1. Approximating Definite Integrals For given integral bf x dx and a n (must be even for Simpson’s Rule) define Dx = b-a and òa ( ) n divide [ ] into n subintervals x [x0, 1] [ 1 …2] x [ n-1,xn] with x0= a and x =nb then, b * * * * Midpoint Rule : òa f ( )x » Dx fºx ( 1 ( )L+ f2x ( n ß, xiis midpoint x[,i-1 i] Trapezoid Rule : b f ( )x » Dx Ø f ( )2 f x ( )f x + ( ) L +2 f ( )+ f ( ) òa 2 º 0 1 2 n-1 n ß b D x Simpson’s Rule : ò f ( )x » º f ( 0 f x +( 1x +2 ( 2 L + f x ( n-2)+ f x( n-1)+ f ( n ß a 3 Visitttp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

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