Calculus Integrals Reduced Cheat Sheet

Calculus Cheat Sheet Calculus Cheat Sheet Integrals Definitions Definite Integral: Suppose f  x is continuous Anti-Derivative : An anti-derivative of f  x  Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. g  b b a f  g  x  g  x  dx   g a f  u  du on a, b . Divide a, b into n subintervals of is a function, F  x , such that F   x   f  x  . u Substitution : The substitution u  g  x  will convert width  x and choose x i* from each interval. Indefinite Integral :  f  x dx  F  x  c du  g   x  dx . For indefinite integrals drop the limits of integration. Then n b where F  x is an anti-derivative of f  x  . f x  x . a f  x  dx  nlim  i  * i 1 Fundamental Theorem of Calculus Variants of Part I : d u x f t  dt  u  x  f u  x  dx a d b d x f  t  dt   v x  f  v x  and g   x   a f t  dt  f  x  . dx v x dx d u x f x is continuous on a , b , F x is Part II :       f  t  dt  u  x  f  u ( x )  v x  f v( x)  dx v x an anti-derivative of f  x  (i.e. F  x   f  x dx )  x is continuous on a, b then x g  x    f  t  dt is also continuous on a, b a then  f  a x dx  F b  F a .  f x   g  x dx   f  x  dx   g x  dx b b b a f  x   g  x  dx  a f  x  dx  a g  x  dx Properties  cf x  dx  c f  x  dx , c is a constant b b  a cf x dx  c  a f x dx , c is a constant a b  c dx  c b  a   f  x  dx   f  x  dx a f  x  dx  0 a a f  x  dx   b f  x  dx b c b a a c b b a a  f x dx   f x dx   f x dx for any value of c. If f x  g  x  on a  x  b then  f  x dx   g  x dx If f  x   0 on a  x  b then b b a a b  f  x dx  0 a b If m  f  x  M on a  x  b then m b  a   f  x dx  M  b  a a Common Integrals k dx  k x  c n n 1  x dx n 1 x  c, n  1 x dx dx    x  ln x  c  a x b dx  a ln ax  b  c ln u du  u ln u  u  c  e du  e  c 1 1 1 1 u 1 u 2 3 cos  x  dx 2 1 5 x 2 cos  x3  dx    cos u du  sin u  c  sin u du   cos u  c  sec u du  tan u  c  sec u tan u du  sec u  c  csc u cot udu   csc u  c  csc u du   cot u  c 2 2 tan u du  ln sec u  c sec u du  ln sec u  tan u  c  a u du  a tan  ua   c  a 1u du  sin  au   c 1 2 1 1 2 1 2 2 x  1  u  13  1 :: x  2  u  2 3  8 Integration by Parts :  u dv  uv   v du and 85 cos 1 3  sin u  1  8 5 3 b b a u dv  uv a 5 3 u  du  sin  8  sin 1  b   v du . Choose u and dv from a integral and compute du by differentiating u and compute v using v   dv . Ex. x e x dx u x Ex. dv  e x  du  dx v  e  x  x e x dx  x e x   e x dx  x e x  e x  c      5 3 ln x dx u  lnx 5 3 dv  dx  du  1x dx v  x ln x dx  x ln x 3   dx   x ln  x   x 5 5 5 3 3 5ln 5  3ln 3  2 Products and (some) Quotients of Trig Functions n m n m For  sin x cos x dx we have the following : For tan xsec 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 convert the rest to secants using cosines using sin2 x  1  cos2 x , then use the substitution u  cos x . tan 2 x  sec2 x 1 , then use the substitution u  sec x . 2. m odd. Strip 1 cosine out and convert rest 2 2 2. m even. Strip 2 secants out and convert rest to sines using cos x  1  sin x , then use to tangents using sec 2 x  1 tan 2 x , then the substitution u  sin x . 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. 4. n even and m odd. Each integral will be and/or half angle formulas to reduce the integral into a form that can be integrated. dealt with differently. Trig Formulas : sin  2x   2sin  x  cos  x  , cos2  x  12 1  cos 2 x  , sin2  x   21  1 cos 2x  a b 2 1 5 x u  x 3  du  3x 2dx  x 2dx  13 du Part I : If f b Ex. using Ex. tan 3 xsec 5 x dx  tan 3 Ex. x sec xdx   tan x sec x tan x sec xdx 5 2  sec x  1 sec   u  1 u du  2 2 4 4 x tan x sec xdx 4  17 sec7 x  15 sec5 x  c u  sec x  sin5  cos xx dx 3 sin5 x (sin 2 x )2 sin x sin4 x sin x  cos x dx   cos x dx   cos x dx   (1cos x) sin x dx  u  cos x cos x   (1 u ) du   1 2 u  u du u u 3 3 3 2 2 3 2 2 3 2 3 4 2 2  12 sec x  2 ln cosx  12 cos x c Calculus Cheat Sheet 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. a2  b2 x2  x  basin  b 2 x 2  a 2  x  ab sec a2  b2 x2  x  ab tan  cos 2 1 sin 2  Ex.  16 x 2 4 9 x 2 tan 2   sec 2   1   dx x  23 sin   dx  23 cos  d sec2   1  tan 2 16 4 2  9 sin  2cos  23 cos   d   sin122  d 4  9 x  4  4 sin 2  4 cos 2   2 cos  Use Right Triangle Trig to go back to x’s. From 3x Recall x 2  x . 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  ’s and remove absolute value bars based on that and,  x if x 0 x   x if x  0 From this we see that cot   x In this case we have 4  9 x 2  2 cos  . Partial Fractions : If integrating 16 2 4 9 x 2 dx  2 4 9x 3x 2  4 4 x9 x P x  Q x dx where the degree of P x  is smaller than the degree 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. Term in P.F.D Factor in Q  x  A ax  b ax  b Ax B 2 ax bx c ax 2  bx  c Ex.   7 x2 13 x ( x 1)( x 2 4) 7 x2 13x ( x 1 )( x2 4 ) dx    ax  b k ax 2 4 3x 16 x  1 x2 4 dx 4  x32 x x1 4 Term in P.F.D A1 A2 Ak    k ax  b ax  b  2 ax  b  A1 x  B1 A kx  B k   2 k ax  bx  c ax 2  bx  c  7 x2 13 x ( x1)( x2 4) dx   bx  c  k  16 dx x2 4   4 ln x 1  32 ln  x 2  4   8 tan 1  2x Here is partial fraction form and recombined. C  xA1  Bx  x2 4 represents the net area between f  x and the x-axis with area above x-axis positive and area below x-axis negative. y  f x   A  b a  upper function    lower function dx & x  f  y   A   d c right function     left function  dy 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. . So, c Q  x  . Factor denominator as completely as possible and find the partial fraction decomposition of Factor in Q x Applications of Integrals a f  x  dx Area Between Curves : The general formulas for the two main cases for each are, 2  12 csc d  12 cot   c 2 Net Area : b A b a f x   g x dx A  d c f  y   g  y  dy A c b a f  x  g  x dx  c g  x   f  x dx 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 Cy linders A    outer radius 2   inner radius 2 A  2  radius  width / height   Limits: x /y of right/bot rin g to x/y of left/top ring Limits : x /y of inner c yl. to x/y of outer cyl. Horz. Axis use f  x  , Vert. Axis use f  y  , Horz. Axis use f y  , Vert. Axis use f  x  , 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  g x radius : a  y radius : a  y inner radius : a  g  x  inner radius: a  f  x width : f  y   g  y  width : f  y  g  y A( x 2 4) (Bx C ) (x 1 ) ( x1)( x2 4) Set numerators equal and collect like terms. 7 x2  13 x   A  B  x2   C  B  x  4 A  C Set coefficients equal to get a system and solve to get constants. A B 7 C  B 13 4 A C 0 A4 B3 C  16 An alternate method that sometimes works to find constants. Start with setting numerators equal in previous example : 7 x2  13 x  A  x2  4    Bx  C  x  1 . Chose nice values of x and plug in. For example if x  1 we get 20  5 A which gives A  4 . This won’t always work easily. outer radius :a  f x 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  0 and x  a  0 ) interchange x and y to get appropriate formulas. Calculus Cheat Sheet Work : If a force of F  x  moves an object Average Function Value : The average value of f x  on a  x  b is favg  b in a  x  b , the work done is W  a F x dx b 1 b a a  f  x dx Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, b b b SA   2 y ds (rotate about x-axis) L   ds a SA   2 x ds (rotate about y-axis) a a where ds is dependent upon the form of the function being worked with as follows.   1   ds  1 dy dx dy ds  2 dx 2  dxdt    dy 2 dx if y  f  x , a  x  b ds  dy if x  f  y  , a  y  b ds  r2   ddr  d if r  f    , a    b 2  dt dt if x  f  t  , y  g  t  , a  t  b 2  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 1. 3.  t  f  x dx lim  f  x dx 2. t  a a  c   f  x  dx    f  x  dx      c b   f  x  dx   b lim  f  x  dx t   t f  x  dx provided BOTH integrals are convergent. Discontinuous Integrand 1. Discont. at a:  b b f  x  dx  lim f  x  dx t a 2. Discont. at b : t a 3. Discontinuity at a  c  b : b c  b a f  x dx  lim t b t  f  x dx a b  f  x  dx   f  x  dx   f  x  dx provided both are convergent. a a c Comparison Test for Improper Integrals : If f  x   g  x   0 on a,   then, 1. If  a  f  x  dx conv. then  g  x dx conv. Useful fact : If a  0 then a  a 1 xp   a a 2. If  g  x  dx divg. then  f  x dx divg. dx converges if p  1 and diverges for p  1. Approximating Definite Integrals b For given integral  a f  x  dx and a n (must be even for Simpson’s Rule) define  x  b na and divide a, b  into n subintervals x 0 , x1 , x1 , x2 , … ,  xn  1 , x n with x0  a and x n  b then, Midpoint Rule :  f  x  dx  x  f  x   f x    f x   , b * 1 a b Trapezoid Rule :  f  x dx  a b Simpson’s Rule :  f  x dx  a x * 2 * n xi* is midpoint  xi 1 , xi   f  x0   2 f  x1  2 f  x2   2 f  x n1   f  x n  2 x  f  x0   4 f  x1  2 f  x2    2 f  xn2   4 f  xn1  f  xn   3