Calculus Cheat Sheet

Calculus Cheat Sheet

Integrals Definitions Anti-Derivative : An anti-derivative of f ( x ) Definite Integral: Suppose f ( x ) is continuous on [ a, b] . Divide [ a, b ] into n subintervals of

is a function, F ( x ) , such that F ¢ ( x ) = f ( x ) .

width D x and choose x from each interval.

Indefinite Integral : ò f ( x ) dx = F ( x ) + c

* i

Then

¥

where F ( x ) is an anti-derivative of f ( x ) .

f (x )D x . ò a f ( x ) dx = nlim å i b

®¥

* i

=1

Fundamental Theorem of Calculus Variants of Part I : Part I : If f ( x ) is continuous on [ a, b ] then d u( x) x f ( t ) dt = u ¢ ( x ) f éëu ( x ) ùû g ( x ) = ò f ( t ) dt is also continuous on [ a, b ] dx ò a a d b d x f ( t ) dt = -v¢ ( x ) f éëv ( x ) ùû and g ¢ ( x ) = f t dt = f x . () ( ) ò v( x ) ò dx a dx d u( x) Part II : f ( x ) is continuous on [ a, b ] , F ( x ) is f ( t ) dt = u ¢ ( x ) f [ u ( x ) ] - v¢ ( x ) f [ v ( x ) ] dx ò v( x ) an anti-derivative of f ( x ) (i.e. F ( x ) = ò f ( x ) dx ) b

then ò f ( x ) dx = F ( b ) - F ( a ) . a

ò f ( x ) ± g ( x ) dx = ò f ( x ) dx ± ò g ( x ) dx b b b ò a f ( x ) ± g ( x ) dx = ò a f ( x ) dx ± ò a g ( x ) dx

Properties

ò cf ( x ) dx = c ò f ( x ) dx , c is a constant b b ò a cf ( x ) dx = c ò a f ( x ) dx , c is a constant

a

b

ò a f ( x ) dx = 0 b

ò c dx = c ( b - a ) ò f ( x ) dx £ ò f ( x ) a

a

ò a f ( x ) dx = -òb f ( x ) dx b

c

b

a

a

c

b

b

a

a

dx

ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx for any value of c. If f ( x ) ³ g ( x ) on a £ x £ b then ò f ( x ) dx ³ ò g ( x ) dx If f ( x ) ³ 0 on a £ x £ b then

b

b

a

a

b

ò f ( x ) dx ³ 0 a

b

If m £ f ( x ) £ M on a £ x £ b then m ( b - a ) £ ò f ( x ) dx £ M ( b - a ) a

ò k dx = k x + c n n 1 ò x dx = n+1 x + c, n ¹ -1 ò x dx = ò x dx = ln x + c ò a x + b dx = a ln ax + b + c ò ln u du = u ln ( u ) - u + c ò e du = e + c +1

-1

1

1

u

1

u

Common Integrals ò cos u du = sin u + c

ò sin u du = - cos u + c ò sec u du = tan u + c ò sec u tan u du = sec u + c ò csc u cot udu = - csc u + c ò csc u du = - cot u + c 2

ò tan u du = ln sec u + c ò sec u du = ln sec u + tan u + c u ò a + u du = a tan ( a ) + c u 1 ò a - u du = sin ( a ) + c 1

2

1

2

-1

-1

2

2

2

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins


Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. ( ) ò a f ( g ( x ) ) g ¢ ( x ) dx = ò g (a ) f ( u ) du b

u Substitution : The substitution u = g ( x ) will convert

g b

using

du = g ¢ ( x ) dx . For indefinite integrals drop the limits of integration. Ex.

2

ò 1 5x

2

cos ( x3 ) dx

2

ò 1 5x

2

cos ( x3 ) dx = ò

u = x 3 Þ du = 3x 2 dx Þ x 2 dx = 13 du

5 cos 1 3 8

= 53 sin ( u ) 1 =

x = 1 Þ u = 1 = 1 :: x = 2 Þ u = 2 = 8 3

8

3

Integration by Parts : ò u dv = uv - ò v du and

b

ò a u dv = uv

b a

5 3

( u ) du

( sin (8) - sin (1) )

b

- ò v du . Choose u and dv from a

integral and compute du by differentiating u and compute v using v = ò dv . Ex.

ò xe

u=x

ò xe

-x

-x

dx

Ex.

dv = e- x Þ

du = dx v = -e - x

dx = - xe + ò e dx = - xe - e -x

-x

-x

-x

+c

5

ò3 ln x dx

u = ln x 5

ò3

dv = dx Þ du = 1x dx v = x

ln x dx = x ln x 3 - ò dx = ( x ln ( x ) - x ) 5

5

5

3

3

= 5ln ( 5) - 3ln ( 3) - 2 Products and (some) Quotients of Trig Functions For ò sin n x cos m x dx we have the following : For ò tan n x sec m x dx we have the following : 1. n odd. Strip 1 sine out and convert rest to 1. 2 2 cosines using sin x = 1 - cos x , then use the substitution u = cos x . 2. m odd. Strip 1 cosine out and convert rest 2. to sines using cos 2 x = 1 - sin 2 x , then use the substitution u = sin x . 3. n and m both odd. Use either 1. or 2. 4. n and m both even. Use double angle 3. and/or half angle formulas to reduce the 4. integral into a form that can be integrated. Trig Formulas : sin ( 2 x ) = 2sin ( x ) cos ( x ) , cos 2 ( x ) =

Ex. ò tan 3 x sec5 x dx

ò tan

3

sin5 x

ò cos x dx (sin x ) sin x sin x sin x sin x ò cos x dx = ò cos x dx = ò cos x dx (1- cos x ) sin x =ò dx ( u = cos x ) cos x = - ò (1-u ) du = - ò 1-2u +u du u u

Ex.

x sec5 xdx = ò tan 2 x sec 4 x tan x sec xdx = ò ( sec2 x - 1) sec 4 x tan x sec xdx = ò ( u 2 - 1) u 4 du

n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using tan 2 x = sec 2 x - 1 , then use the substitution u = sec x . m even. Strip 2 secants out and convert rest to tangents using sec2 x = 1 + tan 2 x , then use the substitution u = tan x . n odd and m even. Use either 1. or 2. n even and m odd. Each integral will be dealt with differently. 2 1 1 2 (1 + cos ( 2 x ) ) , sin ( x ) = 2 (1 - cos ( 2 x ) )

( u = sec x )

= 17 sec7 x - 15 sec5 x + c


3

5

3

2

4

3

2

3

2

2

3

2 2

2

3

3

4

= 12 sec2 x + 2 ln cos x - 12 cos 2 x + c © 2005 Paul Dawkins


Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions. a 2 - b 2 x 2 Þ x = ab sin q

b 2 x 2 - a 2 Þ x = ba sec q

cos 2 q = 1 - sin 2 q Ex.

òx

16 2

4 -9 x 2

tan 2 q = sec 2 q - 1 ó õ

dx

x = 23 sin q Þ dx = 23 cos q dq

sec2 q = 1 + tan 2 q

16 4 sin 2 q ( 2cosq ) 9

Recall x 2 = x . Because we have an indefinite 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 ³ 0 x =í î- x if x < 0

Use Right Triangle Trig to go back to x’s. From substitution we have sin q = 32x so,

From this we see that cot q =

òx

4 - 9x = 2 cos q . 2

Partial Fractions : If integrating

( 23 cos q ) dq = ò sin122 q dq = ò 12 csc 2 dq = -12 cot q + c

2 2 4 - 9x 2 = 4 - 4sin q = 4 cos q = 2 cos q

In this case we have

a 2 + b 2 x 2 Þ x = ab tan q

16 2

4 -9 x 2

dx = - 4

4 -9 x 2 3x 4 -9 x 2 x

. So,

+c

P( x )

ò Q( x) dx where the degree of P ( x ) is smaller than the degree of

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 )

Ex.

ò

ò

Term in P.F.D Factor in Q ( x )

ax + b

A ax + b

ax 2 + bx + c

Ax + B 2 ax + bx + c

( ax + b )

( ax

2

( x -1)( x

7 x2 +13 x

( x -1)( x2 + 4 )

2

Ak x + Bk A1 x + B1 +L + k 2 ax + bx + c ( ax 2 + bx + c )

k

7 x2 +13 x

dx +4)

2

( x -1)( x + 4 )

dx = ò x4-1 + 3xx2++164 dx

= ò x4-1 +

Ak A1 A2 + +L + 2 k ax + b ( ax + b ) ( ax + b )

k

+ bx + c )

7 x2 +13 x

Term in P.F.D

3x x2 + 4

+

16 x2 + 4

dx

= 4 ln x - 1 + 23 ln ( x 2 + 4 ) + 8 tan -1 ( x2 ) Here is partial fraction form and recombined.

=

A

x -1

+C + Bx = x2 + 4

A( x2 + 4) + ( Bx + C ) ( x -1) ( x -1)( x 2 + 4 )

Set numerators equal and collect like terms. 7 x 2 + 13x = ( A + B ) x 2 + ( C - B ) x + 4 A - C Set coefficients equal to get a system and solve to get constants. A+ B = 7 C - B = 13 4A - 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 x 2 + 13x = A ( x 2 + 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.



Applications of Integrals Net Area :

b

ò a f ( x ) dx represents the net area between f ( x ) and the

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, y = f ( x) Þ A = ò

b a

éupper function ù ë û

- éëlower

function ùû dx

& x = f ( y) Þ A = ò

d c

é right function ù ë û

- éëleft

function ùû 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.

b

A = ò f ( x ) - g ( x ) dx a

d

A = ò f ( y ) - g ( y ) dy

c

b

a

c

A = ò f ( x ) - g ( x ) dx + ò g ( x ) - f ( x ) dx

c

Volumes of Revolution : The two main formulas are V = ò A ( x ) dx and V = ò A ( y ) dy . Here is some general information about each method of computing and some examples. Rings Cylinders 2 2 A = 2p ( radius ) ( width / height ) A = p ( outer radius ) - ( inner radius)

(

)

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 ( 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 - f ( x )

outer radius: a + g ( x )

radius : a + y

inner radius : a - g ( x )

inner radius: a + f ( x )

radius : a - y width : f ( y ) - g ( y )

width : f ( y ) - 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 > 0 and x = a £ 0 ) interchange x and y to get appropriate formulas. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.



Work : If a force of F ( x ) moves an object

Average Function Value : The average value of f ( x ) on a £ x £ b is f avg =

b

in a £ x £ b , the work done is W = ò F ( x ) dx a

b 1 b-a a

ò f ( x ) dx

Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, b

b

L = ò ds

b

SA = ò 2p y ds (rotate about x-axis)

a

SA = ò 2p 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

ds =

dx dy

( dxdt )

( )

2

dx if y = f ( x ) , a £ x £ b

ds =

2

dy if x = f ( y ) , a £ y £ b

ds = r 2 + ( ddrq ) dq if r = f (q ) , a £ q £ b

2

+

dy dt

2

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.

ò

¥

t

f ( x ) dx = lim ò f ( x ) dx t ®¥

a

2.

a

¥

c

¥

-

-

c

ò ¥ f ( x ) dx = ò ¥ f ( x ) dx + ò

b

f ( x ) dx = lim

ò¥ -

t ®-¥

b

ò f ( x ) dx t

f ( x ) dx provided BOTH integrals are convergent.

Discontinuous Integrand b

b

b

1. Discont. at a: ò f ( x ) dx = lim+ ò f ( x ) dx a

t ®a

3. Discontinuity at a < c < b :

t

2. Discont. at b : ò f ( x ) dx = lim- ò f ( x ) dx

t

a

b

c

b

a

a

c

t ®b

a

ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx provided both are convergent.

Comparison Test for Improper Integrals : If f ( x ) ³ g ( x ) ³ 0 on [ a, ¥ ) then, ¥

¥

1. If ò f ( x ) dx conv. then ò g ( x ) dx conv. a

a

Useful fact : If a > 0 then

¥

òa

1

xp

¥

¥

a

a

2. If ò g ( x ) dx divg. then ò f ( x ) dx divg.

dx converges if p > 1 and diverges for p £ 1 .

Approximating Definite Integrals For given integral

b

ò a f ( x ) dx and a n (must be even for Simpson’s Rule) define Dx = b-na

and

divide [ a, b] into n subintervals [ x0 , x1 ] , [ x1 , x2 ] , … , [ xn -1 , xn ] with x0 = a and xn = b then, Midpoint Rule : Trapezoid Rule : Simpson’s Rule :

ò f ( x ) dx » Dx éë f ( x ) + f ( x ) + L + f ( x )ùû , xi b

* 1

a

* 2

* n

*

is midpoint [ xi -1 , xi ]

Dx ò f ( x ) dx » 2 éë f ( x ) + 2 f ( x ) + +2 f ( x ) + L + 2 f ( x ) + f ( x )ùû b

a

0

1

2

n -1

n

Dx ò f ( x ) dx » 3 éë f ( x ) + 4 f ( x ) + 2 f ( x ) + L + 2 f ( x ) + 4 f ( x ) + f ( x )ùû b

a

0

1


2

n-2

n -1

n
