( ) Ú Ú. Ú Ú, c is a constant Ú Ú Ú Ú Ú. a Ú for any value of c. Ú Ú Ú Ú Ú Ú Ú ( ) =- ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(1)

Integrals

Definitions Definite Integral: Suppose f x

( )

is continuous

on

[ ]

a b, . Divide

[ ]

a b, into n subintervals of width Dx and choose *

i

x from each interval.

Then

( )

* 1 lim _i b a f x dx n®¥_i₌ f x x ¥ =

å

D

ò

. Anti-Derivative : An anti-derivative of f x

( )

is a function, F x

( )

, such that F x¢

( )

= f x

( )

.

Indefinite Integral :

_ò

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

( )

x

( )

a g x =

ò

f t dt is also continuous on

[ ]

a b, and

( )

x

( )

a d g x f t dt f x dx ¢ =

ò

= . Part II : f x

( )

is continuous on

[ ]

a b, , F x

( )

is an anti-derivative of f x

( )

(i.e. F x

( )

=

ò

f x dx

( )

) then b

( )

a f x dx F b= -F a

ò

. Variants of Part I :

( )

u x a d f t dt u x f u x dx

ò

= ¢ éë ùû

( )

b v x d f t dt v x f v x dx

ò

= - ¢ éë ùû

( )

( ) ( )

( )

[ ]

( )

[ ]

( ) u x v x u x v x d f t dt u x f v x f dx

ò

= ¢ - ¢ Properties

( )

f x ±g x dx= f x dx± g x dx

ò

( )

b b b a f x ±g x dx= a f x dx± ag x dx

ò

( )

0 a a f x dx=

ò

( )

b a a f x dx= - b f x dx

ò

( )

cf x dx c f x dx=

ò

, c is a constant

( )

b b acf x dx c= a f x dx

ò

, c is a constant

(

)

b ac dx c b a=

-ò

( )

b b a f x dx £ a f x dx

ò

( )

b c b a f x dx= a f x dx+ c f x dx

ò

for any value of c.

If f x

( )

³g x

( )

ona x b£ £ then b

( )

b

( )

a f x dx³ a g x dx

ò

If f x

( )

³0 on a x b£ £ then b

( )

0 a f x dx³

ò

If m£ f x

( )

£M on a x b£ £ then

(

)

b

( )

(

)

a m b a- £

ò

f x dx M b a£ -Common Integrals k dx k x c= +

ò

1 1 1 , 1 n n n x dx= ₊ x + +c n¹

-ò

1 1 _ln x x dx- = dx= x c+

ò

1 1_ln a a x b+ dx= ax b c+ +

ò

( )

lnu du u= ln u - +u c

ò

u_du₌ u₊_c

ò

e e cosu du=sinu c+

ò

sinu du= -cosu c+

ò

2 sec u du=tanu c+

ò

sec tanu u du=secu c+

ò

csc cotu udu= -cscu c+

ò

2 csc u du= -cotu c+

ò

tanu du=ln secu c+

ò

secu du=ln secu+tanu c+

ò

( )

1 1 1 2 2 tan u a a a u du c -+ = +

ò

( )

1 2 2 1 _sin u a a u du c -- = +

ò

(2)

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

(

( )

)

( )

g b_{( )}( )

( )

a f g x g x dx¢ = g a f u du

ò

using

( )

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

Ex. 2 2

( )

3 1 5 cosx x dx

ò

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

( )

(

( )

)

2 3 2 _{8 5} 3 1 1 8 5 5 3 ₁ 3 5 cos cos

sin sin 8 sin 1

x x dx u du

u

=

= =

-ò

ò

Integration by Parts :

ò

u dv uv= -

ò

v du and b b_a b

au dv uv= - av du

ò

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

_ò

dv.

Ex. _x -x_dx

ò

e x x u x dv₌ ₌_e- _Þ du dx v₌ _{= -}_e -x x x x x x - dx_{= -}x - ₊ - dx_{= -}x - _- - ₊c

ò

e e

ò

e e e Ex. 5 3lnx dx

ò

1 ln _x u= x dv dx= Þ du= dx v x=

( )

(

)

( )

5 ₅ 5 5 3 3 3ln ln 3 ln 5ln 5 3ln 3 2 x dx x= x - dx= x x -x = -

-ò

ò

Products and (some) Quotients of Trig Functions

For sinn_xcosm_{x dx}

ò

we have the following :

1. n odd. Strip 1 sine out and convert rest to cosines using _sin2_x_{= -}_{1 cos}2_x_{, then use}

the substitution u=cosx.

2. m odd. Strip 1 cosine out and convert rest to sines using _cos2_x_{= -}_{1 sin}2_x_{, then use}

the substitution u=sinx.

3. n and m both odd. Use either 1. or 2.

4. n and m both even. Use double angle and/or half angle formulas to reduce the integral into a form that can be integrated.

For tann_xsecm_{x dx}

ò

we have the following :

1. n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using

2 2

tan x=sec x-1, then use the substitution sec

u= x.

2. m even. Strip 2 secants out and convert rest to tangents using _sec2_x_{= +}_{1 tan}2_x_{, then}

use the substitution u=tanx.

3. n odd and m even. Use either 1. or 2.

4. n even and m odd. Each integral will be dealt with differently.

Trig Formulas : sin 2

( )

x =2sin

( ) ( )

x cos x , 2

( )

1

(

( )

)

2

cos x = 1 cos 2+ x , 2

( )

1

(

( )

)

2

sin x = 1 cos 2- x Ex.

ò

_tan3_x_sec5_{x dx}

(

)

(

)

(

)

3 5 2 4 2 4 2 4 7 5 1 1 7 5

tan sec tan sec tan sec

sec 1 sec tan sec

1 sec sec sec x xdx x x x xdx x x x xdx u u du u x x x c = = -= - = = - +

ò

Ex. sin5₃ cos x xdx

ò

(

)

2 2 1 1 2 2 2 2 5 4 3 3 3 2 2 3 2 2 ₂ ₄ 3 3 sin (sin ) sin sin sin

cos cos cos

sin (1 cos )

cos

(1 ) _{1 2}

cos

sec 2ln cos cos

x x x x x x x x x x x u _u _u u u dx dx dx dx u x du du x x x c -- _- ₊ = = = = = - = -= + - +

ò

(3)

Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions.

2 2 2 a_sin b a -b x Þ x= q 2 2 cos q = -1 sin q 2 2 2 a_sec b b x -a Þ x= q 2 2 tan q =sec q -1 2 2 2 a_tan b a +b x Þ x= q 2 2 sec q = +1 tan q Ex. ₂ 2 16 4 9 x - x dx

ò

2 2 3sin 3cos x= q Þ dx= q qd 2 2

2 _{4 4sin} _4cos _{2 cos}

4 9- x = - q = q = q

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 q’s and remove absolute value bars based on that and,

if 0 if 0 x x x x x ³ ì = í_- _< î

In this case we have _{4 9}_- _x2 ₌_2cos_q _.

( )

(

)

2 3 sin 2cos 2 2 2 4 9 16 12 sin cos 12csc 12cot d d d c q q q q q q q q = = = - + ó õ

ò

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

From this we see that cotq ₌ 4 9₃-_xx2 _{. So,}

2 2 2 16 4 4 9 4 9 x x x x dx c -- = - +

ò

Partial Fractions : If integrating

ò

_{Q x}P 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

( )

Term in P.F.D Factor in Q x

( )

Term in P.F.D

ax b+ A ax b+

(

)

k ax b+ 1

₍

2

₎

2

₍

₎

k k A A A ax b+ + ax b+ + +L ax b+ 2 ax +bx c+ ₂Ax B ax bx c + + +

(

)

2 k ax +bx c+ 21 1

₍

₂

₎

k k k A x B A x B ax bx c _ax _{bx c} + + _{+ +} + + L ₊ ₊ Ex. ₍ ₎₍ 2 ₎ 2 1 4 7 13 x x x x _dx - + +

ò

(

)

( )

2 2 2 2 ( )( ) 2 1 3 2 2 2 ₄ _{3 16} 1 1 4 4 3 16 4 1 ₄ ₄ 7 13 4ln 1 ln 4 8 tan x x x x x x x _x _x x x x dx dx dx x x -+ -- + + - ₊ ₊ + ₌ ₊ = + + = - + + +

ò

Here is partial fraction form and recombined.

2 2 2 2 4) ( ) ( ) ( )( ) ( )( ) 2 1 1 1 4 4 1 4 ( 7 13 Bx C x x x x x x x A x Bx C A x x + + + -- + + - + + + ₌ ₊ ₌

Set numerators equal and collect like terms.

(

)

(

)

2 2

7x +13x= A B x+ + C B x- +4A C

-Set coefficients equal to get a system and solve to get constants. 7 13 4 0 4 3 16 A B C B A C A B C + = - = - = = = =

An alternate method that sometimes works to find constants. Start with setting numerators equal in previous example : ₇_x2₊₁₃_{x A x}₌

(

2 _{+ +}₄

)

(

_{Bx C}₊

) (

_x_-₁

)

_{. Chose}_nice_{values of}_x_{and plug in.}

(4)

Applications of Integrals

Net Area : b

( )

a f x dx

ò

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 upper function lower function

a

y= f x Þ A=

ò

_ëé _ûù-_ëé _ûùdx &

( )

d right function left function

c

x= f y Þ A=

ò

é_ë _ûù-é_ë ù_ûdy

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 A=

_ò

f x -g x dx A=

ò

_cd f y

( ) ( )

-g y dy c

_{( ) ( )}

b

_{( )}

a c A=

ò

f x -g x dx+

ò

g x - f x dx

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

)

outer radius inner radius

A=p - A=2p(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

( )

,

( )

g x ,A x

( )

and dx.

Vert. Axis use f y

( )

,

( )

g y ,A y

( )

and dy.

Horz. Axis use f y

( )

,

( )

g y ,A y

( )

and dy.

Vert. Axis use f x

( )

,

( )

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-

( )

inner radius : a g x-

( )

outer radius: a +g x

( )

inner radius: a + f x

( )

radius :a y -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 0

y a= £ case with a=0. For vertical axis of rotation (x a= >0 and x a= £0) interchange x and

(5)

Work : If a force ofF x

( )

moves an object ina x b£ £ , the work done is b

( )

a

W =

ò

F x dx

Average Function Value : The average value of f x

( )

on a x b£ £ is _avg 1 b

( )

a

b a

f = _-

ò

f x dx

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

b a

L=

_ò

ds b2

a

SA=

_ò

py ds (rotate about x-axis) b2

a

SA=

_ò

px ds (rotate about y-axis) where ds is dependent upon the form of the function being worked with as follows.

( )

2

( )

1 dy_dx if , ds= + dx y= f x a x b£ £

( )

2

( )

1 dx_dy if , ds= + dy x= f y a£ £y b

( )

2

( )

2

( )

if , , dy dx dt dt ds= + dt x= f t y g t= a t b£ £

( )

2

( )

2 dr _if _, d ds= r + _q dq r= f q a£ £q 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.

( )

lim t

( )

a f x dx t®¥ a f x dx ¥ =

ò

2. b

( )

lim b

( )

t t f x dx f x dx -¥ = ®-¥

ò

3.

( )

c

( )

c f x dx f x dx f x dx - -¥ ¥ ¥ = ¥ +

ò

provided BOTH integrals are convergent.

Discontinuous Integrand 1. Discont. at a: b

( )

lim b

( )

a f x dx=t®a+ t f x dx

ò

2. Discont. at b : b

( )

lim t

( )

a f x dx=t b® - a f x dx

ò

3. Discontinuity at a c b< < : b

( )

c

( )

b

( )

a f x dx= a f x dx+ c f x dx

ò

provided both are convergent.

Comparison Test for Improper Integrals : If f x

( )

³g x

( )

³0on

[

a,¥

)

then,

1. If

( )

a f x dx ¥

ò

conv. then

( )

a g x dx ¥

ò

conv. 2. If

( )

a g x dx ¥

ò

divg. then

( )

a f x dx ¥

ò

divg.

Useful fact : If a>0 then 1

a xpdx ¥

ò

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

Approximating Definite Integrals

For given integral b

( )

a f x dx

ò

and a n (must be even for Simpson’s Rule) define D =x b a_n- and divide

[ ]

a b, into n subintervals

[

x x₀, ₁

]

,

[

x x₁, ₂

]

, … ,

[

x_n_-₁,x_n

]

with x₀ =a and x_n =b then,

Midpoint Rule :

( )

( ) ( )

* *

( )

* 1 2 b n a f x dx» Dx f xéë + f x + + f x ùû

ò

L , * i x is midpoint

[

x_i_-₁,x_i

]

Trapezoid Rule :

( )

₀ 2

( )

₁ 2

( )

₂ 2

( )

₁

( )

2 b n n a x f x dx»D é_ëf x + f x + + f x + + f x _- + f x ù_û

ò

L Simpson’s Rule :

( )

₀ 4

( )

₁ 2

( )

₂ 2

(

₂

)

4

( )

₁

( )

3 b n n n a x f x dx»D _ëéf x + f x + f x + + f x_- + f x _- + f x ù_û

ò

L