Calculus Cheat Sheet Integrals - Lamar University

Calculus Cheat Sheet

Integrals

Definitions

Definite Integral: Suppose f ( x) is continuous Anti-Derivative : An anti-derivative of f ( x)

on [a,b] . Divide [a,b] into n subintervals of

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

width x and choose xi* from each interval.

( ) Then= b f ( x) dx a

n

lim f

n i=1

xi* x .

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

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

Fundamental Theorem of Calculus

Part I : If f ( x) is continuous on [a,b] then

g

(

x

)

=

x a

f

(t ) dt

is also continuous on

[a,b]

= and g( x) dd= x ax f (t ) dt f ( x) .

Part II : f ( x) is continuous on[a,b] , F ( x) is

an anti-derivative of f ( x) (i.e. F ( x) = f ( x) dx )

Variants of Part I :

d dx

u(x)

a

f

(t ) dt

=

u( x)

f

u ( x)

db

dx v(x)

f

(t ) dt

=

-v( x)

f

v ( x)

d = u(x) f (t ) dt

dx v(x)

u( x) f [u(x)] - v( x) f [v(x)]

then

b

a

f

(

x= ) dx

F (b)- F (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

a

f

(

x)

?

g

(

x)

dx=

b

a

f

(

x)

dx

?

b

a

g

(

x)

dx

b

a cf

( x) dx

=

b

ca

f

( x) dx

,

c

is

a

constant

a

a

f

(

x)

dx

=

0

b

a c= dx

c(b - a)

b

a

f

( x) dx =

a

-b

f

( x) dx

b

a

f

( x) dx

b

a

f

(x)

dx

a= b f ( x) dx

c

a

f

(

x

)

dx

+

b

c

f

( x) dx

for

any

value

of

c.

If

f ( x) g ( x) on a x b then

b

a

f

( x) dx

b

a

g

(

x)

dx

If

f

(x)

0

on

a

x

b

then

b

a

f

( x) dx

0

If

m

f

(x)

M

on

a

xb

then

m(b - a)

b

a

f

( x) dx

M

(b - a)

k d=x k x + c

= xn dx

1 n+1

xn+1

+

c,

n

-1

x-1= dx x1= dx ln x + c

Common Integrals

cos u= du sin u + c sin u du = - cos u + c sec2 u= du tan u + c

1 = dx ax +b

1

a

ln

ax

+

b

+

c

ln u= du u ln (u) - u + c

eu du= eu + c

sec u tan u= du sec u + c csc u cot udu = - csc u + c csc2 u du = - cot u + c

Visit for a complete set of Calculus notes.

ta= n u du ln secu + c

sec u du= ln sec u + tan u + c

( ) = a2+1 u2 du

1

a

tan -1

u a

+c

( ) = 1 du a2- u2

sin -1

u a

+c

? 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) will convert

b

a

f

( g ( x)) g( x) dx

=

g(b)

g(a)

f

(u)

du

using

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

( ) Ex. 2 5x2 cos x3 dx 1

u =x3

du =3x2dx

x2dx

=

1 3

du

x =1 u =13 =1 :: x = 2 u = 23 = 8

( ) 25x2 cos 1

x3

dx =

8 1

5 3

cos

(u

)

du

=

5 3

s= in (u) 8 1

5 3

(sin

(8)

-

sin

(1)

)

Integration by Parts : u d=v

b

uv - v du and a u= dv

uv

b a

-

b

v du .

a

Choose u and dv from

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

Ex. xe-x dx

u = x dv = e-x du = dx v = -e-x

xe-x dx = -xe-x + e-x dx = -xe-x - e-x + c

5

Ex. 3 ln x dx

u = ln x

dv = dx

du

=

1 x

dx

v =x

( ) 5

ln x dx =

3

x

ln

x

5 3

-

5

dx =

3

x ln ( x) - x 5 3

= 5ln (5) - 3ln (3) - 2

Products and (some) Quotients of Trig Functions

For sinn x cosm x dx we have the following :

For tann x secm 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 sin2 x= 1- cos2 x , then use the substitution u = cos x .

convert the rest to secants using ta= n2 x sec2 x -1, then use the substitution

2. m odd. Strip 1 cosine out and convert rest

u = sec x .

to sines using cos2 x= 1- sin2 x , then use the substitution u = sin x .

2. m even. Strip 2 secants out and convert rest to tangents using sec2 x= 1+ tan2 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 (2x) = 2sin ( x) cos ( x) , cos2 (= x)

1 2

(1+

cos(2x)) ,

sin 2

(= x)

1 2

(1

-

cos

(

2

x

)

)

Ex. tan3 x sec5 x dx

Ex.

sin5 x cos3 x

dx

tan3 x sec5 xdx = tan2 x sec4 x tan x sec xdx

= csions53xx dx = sinc4oxs3sixn x dx

(sin2 x)2 sin cos3 x

x

dx

( ) =

sec2 x -1 sec4 x tan x sec x= dx ( ) = (1-cocso2sx3)x2 sin x dx u cos x

( ) = u2 -1 u4du

(u = sec x)

=

1 7

sec7

x

-

1 5

sec5

x+c

= - (1-uu32 )2 du = - 1-2uu23+u4 du

=

1 2

sec2

x

+

2 ln

cos

x

-

1 2

cos2

x

+

c

Visit 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.

a2 - b2 x2 x = ba sin cos2 = 1- sin2

b2 x2 - a2 x = ba sec ta= n2 sec2 -1

a2 + b2 x2 x = ba tan sec2 = 1+ tan2

Ex.

16 dx

x2 4-9x2

=x

2 3

sin

d=x

2 3

cos

d

4 - 9x2 =4 - 4sin2 =4 cos2 = 2 cos

Recall x2 = 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 's and remove absolute value bars based on that and,

x if x 0 x = -x if x < 0

In this case we have 4 - 9x2 = 2 cos .

( ) 16

4 9

sin2

(

2

cos

)

2 3

cos

d =

12 sin2

d

= 12 csc2 d = -12 cot + c

Use Right Triangle Trig to go back to x's. From

substitution we have

sin

=

3x 2

so,

From this we see that cot = 4-9x2 . So,

3x

x2

16 4-9 x2

dx

= - 4 4-x9x2

+c

Partial Fractions : If integrating

P(x) dx

Q(x)

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

ax + b

A ax + b

(ax + b)k

ax2 + bx + c

( ) Ax + B

ax2 + bx + c

ax2 + bx + c k

Term in P.F.D

A1 ax + b

+

A2

(ax + b)2

+ +

Ak

(ax + b)k

( ) A1x

ax2 +

+ B1 bx +

c

+

+

Ak x + Bk ax2 + bx + c k

Ex.

7 x2 +13x ( x-1)( x2 +4)

dx

= + = 7x2 +13x

( x-1)( x2 +4)

A Bx+C x-1 x2 +4

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

(

7 x2 x-1)

(+x123+x4)= dx

4 x-1

+

3x+16 x2 +4

dx

=

4 x-1

+

3x x2 +4

+

16 x2 +4

dx

Set numerators equal and collect like terms.

7x2 +13x = ( A + B) x2 + (C - B) x + 4A - C

Set coefficients equal to get a system and solve

( ) =

4 ln

x -1

+

3 2

ln

x2 + 4

+

8

tan

-1

(

x

2

)

to get constants. 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

( ) previous example : 7x2 +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 = 5A which gives A = 4 . This won't always work easily.

Visit for a complete set of Calculus notes.

? 2005 Paul Dawkins

Calculus Cheat Sheet

Applications of Integrals

Net Area :

b

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,

( ) ( ) b

d

=y f x =A a upper - function lower function dx & =x f y =A 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.

=A

b

a

f

(

x)

-

g

(

x)

dx

=A

d

c

f

( y) -

g

( y) dy

A

=

c

a

f

( x) -

g

( x) dx

+

b

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

Cylinders

( ) =A

- ( ) ( ) 2

2

outer radius

inner radius

A = 2 (radius) (width / 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 ( 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) width : f ( y) - g ( y)

radius : a + 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 for a complete set of Calculus notes.

? 2005 Paul Dawkins

Calculus Cheat Sheet

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

in a

x

b

,

the

work

done

is

W

=

b

a

F

( x) dx

Average Function Value : The average value

of

f

(x)

on

a

xb

is

favg

=

1 b-a

b f ( x) dx

a

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

b

L = a ds

b

SA = a 2 y ds (rotate about x-axis)

b

SA = a 2 x ds (rotate about y-axis)

where ds is dependent upon the form of the function being worked with as follows.

( ) ( ) ( ) ds =

1 +

dy

2

dx

if y =

f (x), a x b

= ds

dx

dx

2

+

dy

2

dt

= if x

dt

dt

f (t= ), y

g (t), a t b

( ) ( ) ds =

1 +

dx

2

dy

dy

if x =

f (y), a y b

ds=

r2 +

dr d

2 d

if r=

f ( ), a b

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. f ( x) dx = lim t f ( x) dx

a

t a

2. b f ( x) dx = lim b f ( x) dx

-

t- t

3.

= - f ( x) dx

c

-

f

(

x

)

dx

+

c

f

( x) dx

provided

BOTH

integrals

are

convergent.

Discontinuous Integrand

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

a

ta+ t

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

a

tb- a

3.

Discontinuity at a < c < b : = ab f ( x) dx

c

a

f

( x) dx

+

b

c

f

( x) dx

provided

both

are

convergent.

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

1.

If a

f

( x) dx

conv.

then

a

g

(

x

)

dx

conv.

2.

If

a

g

(

x

)

dx

divg.

then a

f

( x) dx

divg.

Useful fact : If a > 0 then

a

1

xp

dx

converges if

p > 1 and diverges for

p 1.

Approximating Definite Integrals

For given integral

b a

f

( x) dx

and a n (must be even for Simpson's Rule) define

x =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 :

b a

f

( x) dx

x

f

x1* + f

x2*

+ +f

[ ] xn* , xi* is midpoint xi-1, xi

Trapezoid

Rule

:

b

a

f

( x) dx

x 2

f

( x0 ) + 2 f

( x1 ) + +2 f

( x2 ) + + 2 f

( xn-1 ) +

f

( xn )

Simpson's Rule :

b a

f

( x) dx

x 3

f

( x0 ) +

4

f

( x1 ) +

2

f

( x2 ) + + 2

f

( xn-2 ) + 4

f

( xn-1 ) +

f

( xn )

Visit for a complete set of Calculus notes.

? 2005 Paul Dawkins

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