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Calculus Cheat Sheet Integrals Definitions Definite Integral: Suppose Fx( ) Is Continuous Anti-Derivative : an Anti-Derivative of Fx( ) on [Ab, ]

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Calculus Cheat Sheet Integrals Definitions Definite Integral: Suppose Fx( ) Is Continuous Anti-Derivative : an Anti-Derivative of Fx( ) on [Ab, ] Calculus Cheat Sheet Integrals Definitions Definite Integral: Suppose fx( ) is continuous Anti-Derivative : An anti-derivative of fx( ) on [ab, ] . Divide [ab, ] into n subintervals of is a function, Fx( ) , such that Fx′( ) = fx( ) . width ∆ x and choose x* from each interval. = + i Indefinite Integral : ∫ f( x) dx F( x) c b n Then f( x) dx= lim f x* ∆ x . where Fx( ) is an anti-derivative of fx( ) . ∫a n→∞ ∑ ( i ) i=1 Fundamental Theorem of Calculus Part I : If fx( ) is continuous on [ab, ] then Variants of Part I : d ux( ) x f( t) dt= u′( x) f u( x) g( x) = f( t) dt is also continuous on [ab, ] ∫a ∫a dx d x d b and g′( x) = f( t) dt= f( x) . ∫ f( t) dt= − v′( x) f v( x) dx ∫a dx vx( ) d ux( ) Part II : fx( ) is continuous on[ab, ] , Fx( ) is ftdtuxf( ) = ′′( ) [ux()]− vxf( ) [vx()] dx ∫vx( ) an anti-derivative of fx( ) (i.e. F( x) = ∫ f( x) dx ) b then f( x) dx= F( b) − F( a) . ∫a Properties ∫f( x) ±= g( x) dx ∫∫ f( x) dx ± g( x) dx ∫∫cf( x) dx= c f( x) dx , c is a constant b bb bb f( x) ±= g( x) dx f( x) dx ± g( x) dx cf( x) dx= c f( x) dx , c is a constant ∫a ∫∫ aa ∫∫aa a b f( x) dx = 0 c dx= c( b − a) ∫a ∫a ba bb f( x) dx= − f( x) dx f( x) dx≤ f( x) dx ∫∫ab ∫∫aa b cb f( x) dx= f( x) dx + f( x) dx for any value of c. ∫∫∫a ac bb If f( x) ≥ gx( ) on axb≤≤then f( x) dx≥ g( x) dx ∫∫aa b If fx( ) ≥ 0 on axb≤≤ then f( x) dx ≥ 0 ∫a b If m≤≤ fx( ) Mon axb≤≤ then mba( −≤) fxdxMba( ) ≤( −) ∫a Common Integrals ∫ k dx= k x + c ∫ cosu du= sin u + c ∫ tanu du= ln sec u + c nn=1 +1 + ≠− =−+ = ++ ∫ x dxn+1 x c,1 n ∫sinu du cos u c ∫secu du ln sec u tan u c x−1 dx=1 dx =ln x + c sec2 u du= tan u + c 1 du =1 tan −1 ( u ) + c ∫∫x ∫ ∫ au22+ aa 11= ++ = + 1 −1 u ∫ + dxa ln ax b c ∫secu tan u du sec u c du =sin ( ) + c ax b ∫ au22− a ∫ lnu du= u ln ( u) −+ u c ∫ cscu cot udu=−+ csc u c 2 ∫eeuudu= + c ∫ cscu du=−+ cot u c Visit http://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. b gb( ) u Substitution : The substitution u= gx( ) will convert f( g( x)) g′( x) dx= f( u) du using ∫∫a ga( ) du= g′( x) dx . For indefinite integrals drop the limits of integration. 2 28 Ex. 5x23 cos x dx 5x23 cos x dx= 5 cos( u) du ∫1 ( ) ∫∫11( ) 3 =⇒=3 22 ⇒ =1 558 u x du3 x dx x dx3 du =sinu = sin 8 − sin 1 33( ) 1 ( ( ) ( )) 33 xu=⇒==111::228 xu =⇒== bbb Integration by Parts : u dv= uv − v du and u dv= uv − v du . Choose u and dv from ∫∫∫∫aaa integral and compute du by differentiating u and compute v using v= ∫ dv . − x 5 Ex. xe dx Ex. ln x dx ∫ ∫3 −−xx u= x dv =⇒=eedu dx v =− 1 u=ln x dv =⇒= dx dux dx v = x −−−−−xxxxx xeeeee dx=−+=−−+ x dx x c 555 5 ∫∫ lnx dx= x ln x −= dx xln ( x) − x ∫∫333 ( ) 3 =−−5ln( 5) 3ln( 3) 2 Products and (some) Quotients of Trig Functions For ∫sinnmx cos x dx we have the following : For ∫ tannmx sec x dx 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 sin22xx= 1 − cos , then use convert the rest to secants using the substitution ux= cos . tan22xx= sec − 1, then use the substitution 2. m odd. Strip 1 cosine out and convert rest ux= sec . to sines using cos22xx= 1 − sin , then use 2. m even. Strip 2 secants out and convert rest 22 the substitution ux= sin . to tangents using secxx= 1 + tan , then 3. n and m both odd. Use either 1. or 2. use the substitution ux= tan . 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. 2 1 2 1 Trig Formulas : sin( 2x) = 2sin( xx) cos( ) , cos( xx) =2 ( 1 + cos( 2 )), sin( xx) =2 ( 1 − cos( 2 )) 5 Ex. tan35x sec x dx Ex. sin x dx ∫ ∫ cos3 x 35 24 54 22 ∫∫tanx sec xdx= tan x sec x tan x sec xdx sinxdx= sin xx sin dx = (sinx ) sin x dx ∫∫cos33xxcos ∫cos 3 x 24 =secx − 1 sec x tan x sec xdx (1− cos22x ) sin x ∫( ) = dx( u= cos x) ∫ cos3 x 24 =−=(u1) u du( usec x) (1−u22 ) 24 ∫ =−=−du 12−+uudu ∫∫uu33 =−+1175 75secx sec xc 1122 =+22secx 2ln cos x −+ cos xc Visit http://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. 2− 22 ⇒=a θ 22− 2 ⇒=a θ 2+ 22 ⇒=a θ a bx x b sin bx a x b sec a bx x b tan cos22θθ= 1 − sin tan22θθ= sec − 1 sec22θθ= 1 + tan Ex. 16 dx ⌠ 16 2 12 ∫ 2 2 ( 3 cosθθ)dd= 2 θ x 49− x ⌡ 4 sin2θθ( 2cos ) ∫ sin θ 9 x =22sinθ ⇒=dxcosθθ d 33 =∫12csc2 dcθθ =−+ 12cot 2 22 49− x =−==4 4sinθ 4cos θθ 2cos Use Right Triangle Trig to go back to x’s. From 2 θ = 3x Recall xx= . Because we have an indefinite substitution we have sin 2 so, integral we’ll assume positive and drop absolute value bars. If we had a definite integral we’d need to compute θ ’s and remove absolute value bars based on that and, xxif ≥ 0 − 2 x = From this we see that cotθ = 49x . So, −<xxif 0 3x 16 449− x2 2 dx =−+c In this case we have 49− x = 2cosθ . ∫ x2 49− x2 x Px( ) Partial Fractions : If integrating dx where the degree of Px( ) is smaller than the degree of ∫ Qx( ) Qx( ) . 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 Qx( ) Term in P.F.D Factor in Qx( ) Term in P.F.D AA A A k 12+ ++ k ax+ b (ax+ b) 2 k ax+ b ax+ b (ax++ b) ( ax b) Ax11+ B Axkk+ B Ax+ B 2 k ++ 2 2 k ax++ bx c 2 (ax++ bx c) 2 ax++ bx c ax++ bx c (ax++ bx c) 2 2 2 ++ +− Ex. 7xx+ 13 dx 7xx+ 13 =+=A Bx+ C Ax( 4) (Bx C ) ( x 1 ) ∫ (xx−+14 )(2 ) (xx−+14 )(22 ) x−1 x + 4( xx −+ 14 )( 2 ) 2 7xx+ 13 dx =4 + 3x+ 16 dx Set numerators equal and collect like terms. ∫∫(xx−+14 )(2 ) x−1 x2 +4 7x22+ 13 x =+( A Bx) +−( C Bx) +−4 A C =4 ++3x 16 dx ∫ x−1 xx22++44 Set coefficients equal to get a system and solve 3 21− x to get constants. =4lnxx −+ 122 ln( + 4) + 8 tan ( ) AB+=7 C −= B13 4 AC −= 0 Here is partial fraction form and recombined. ABC= 4 = 3 =16 An alternate method that sometimes works to find constants. Start with setting numerators equal in previous example : 7x22+ 13 x = 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. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Applications of Integrals b Net Area : f( x) dx represents the net area between fx( ) and the ∫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( x) ⇒= A upper function − lower function dx & x= f( y) ⇒= A right function − left function dy ∫a ∫c 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 d cb A= f( x) − g( x) dx A=∫ f( y) − g( y) dy A=−+− f( x) g( x) dx g( x) f( x) dx ∫a c ∫∫ac Volumes of Revolution : The two main formulas are V= ∫ A( x) dx and V= ∫ A( y) dy . Here is some general information about each method of computing and some examples. Rings Cylinders 22 A = π ((outer radius) − ( inner radius) ) A = 2π (radius) ( width / height) Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl.
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