INTEGRAL CALCULUS

Integration or

Anti-derivatives

Definition Indefinite Integral

The set of all antiderivatives of a function f(x) is theindefinite integral of f with respect to x and is denoted by

∫ dxxf )(

CxFdxxf +=∫ )()(

Definite Integration

( ) ( ) ( )afbfdxxfb

a

−=∫ '

Antiderivatives with Slope Fields

=∫ dxxn.1 1,1

1

−≠++

+

nCn

xn

=∫ x

dx.2 Cx +ln

=∫ dxex.3 Cex +

=∫ dxxsin.4 Cx +− cos

Antiderivatives with Slope Fields

=∫ dxxcos.5

=∫ dxx2sec.6

=∫ dxx2csc.7

=∫ dxxx tansec.8

Cx +sin

Cx +sec

Cx +− cot

Cx +tan

=∫ dxxx cotcsc.9 Cx +− csc

Antiderivatives with Slope Fields

∫ +− dxxx )32( 3 Cxxx ++− 34

24

∫ dxxx3Cx +2

9

9

2

∫

− dttt

cos12 Ct

t+−− sin

1

Antiderivatives with Slope Fields

∫ + dxxxx )cos(tansec Cxx ++sec

∫ + dxxx )cscsin1( 2 Cxx +− cos

∫ + dtte t )( 22 Cte t

++32

32

Antiderivatives with Slope Fields

Find the position function for v(t) = t3 - 2t2 + t s(0) = 1

Cttt

ts ++−=23

2

4)(

234

C++−=2

)0(

3

)0(2

4

01

234C = 1

123

2

4)(

234

++−= tttts

Anti-derivatives with Slope Fields

=∫ cabin

cabind )(log cabin

log cabin + C houseboat

The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.

Integration by Substitution

( ) 52x dx+∫ Let 2u x= +

du dx=5u du∫61

6u C+

( ) 62

6

xC

++

The variable of integration must match the variable in the expression.

Don’t forget to substitute the value for u back into the problem!

Integration by Substitution

21 2 x x dx+ ⋅∫One of the clues that we look for is if we can find a function and its derivative in the integrand.

The derivative of is .21 x+ 2 x dx1

2 u du∫3

22

3u C+

( )3

2 22

13

x C+ +

2Let 1u x= +2 du x dx=

Note that this only worked because of the 2x in the original.Many integrals can not be done by substitution.

Integration by Substitution

4 1 x dx−∫ Let 4 1u x= −

4 du dx=

1

4du dx=

Solve for dx.

1

21

4

u du⋅∫3

22 1

3 4u C⋅ +

3

21

6u C+ ( )

3

21

4 16

x C− +

Integration by Substitution

( )cos 7 5 x dx+∫7 du dx=

1

7du dx=

1cos

7u du⋅∫

1sin

7u C+

( )1sin 7 5

7x C+ +

Let 7 5u x= +

Integration by Substitution

( )2 3sin x x dx∫ 3Let u x=23 du x dx=

21

3du x dx=

We solve for because we can find it in the integrand.

2 x dx

1sin

3u du∫

1cos

3u C− +

31cos

3x C− +

Integration by Substitution

4sin cos x x dx⋅∫ Let sinu x=

cos du x dx=( ) 4sin cos x x dx∫4 u du∫ 51

5u C+

51sin

5x C+

Integration by Substitution

24

0tan sec x x dx

π

∫The technique is a little different for definite integrals.

Let tanu x=2sec du x dx=

( )0 tan 0 0u = =

tan 14 4

uπ π = =

1

0 u du∫ We can find new

limits, and then we don’t have to substitute back.

new limit

new limit1

2

0

1

2u

1

2We could have substituted back and used the original limits.

Integration by Substitution

24

0tan sec x x dx

π

∫ Let tanu x=2sec du x dx=

4

0 u du

π

∫

Wrong!The limits don’t match!

( ) 42

0

1tan

2x

π

=

( )2

21 1tan tan 0

2 4 2

π = −

2 21 11 0

2 2= ⋅ − ⋅

u du∫21

2u= 1

2=

Using the original limits:

Leave the limits out until you substitute back.

This is usually more work than finding new limits

Integration by Substitution

1 2 3

13 x 1 x dx

−+∫ 3Let 1u x= +

23 du x dx=( )1 0u − =

( )1 2u =12

2

0 u du∫23

2

0

2

3u

Don’t forget to use the new limits.

3

22

23

⋅2

2 23

= ⋅ 4 2

3=

Integration by Substitution

Integration by Substitution

dxxx∫ + 212

dxx∫ + 43

( ) dttt∫ + 8235

dtt∫ tan

θθθ d∫ cossin3

θθθ d∫ 2csccot

Integration by Substitution

dxx

x∫ −

−5

2 1

2

dxxx 2sincos4

2∫

−

−

π

π

xx

dxe

e ln

2

∫

dxe

ex

x

∫ + 5

5

3

∫1

0 xe

dx

dxxx∫ − 21

2 12. Find (1 ) . Let sinx dx u x−− =∫cosdx udu=

2 2(1 ) 1 sin cosx dx u udu− = −∫ ∫2cos u du= ∫

sinx u= Substituting gives,

We can not integrate this yet. Let us use trig.

2 1cos (1 cos2 )

2u u= +

1 1cos2

2 2u du= +∫ 1

sin22 4u

u c= + +

1.2sin cos

2 4u

u u c= + + ( )1sin cos

2u u u c= + +

( )1 21sin 1

2x x x c−= + − +

Now for some trig play…..

sin , but what does cos equal?x u u=

( )1sin cos

2u u u c= + +

2 2sin cos 1u u+ =2 2cos 1 sinu = −

2cos 1 sinu u= −

( )2 1(1 ) sin cos

2x dx u u u c− = + +∫

2

23. Find . Let x 2sin

4

xdx

xθ=

−∫

2cosdx dθ θ=2 2

2 2

4sin .2cos

4 4 4sin

xdx d

x

θ θ θθ

=− −∫ ∫

2

2

4sin .2cos

2 1 sind

θ θ θθ

=−∫

2

2

2sin .2cos

cosd

θ θ θθ

= ∫24sin dθ θ= ∫ 2 1 1

sin cos22 2

θ θ = − ÷

2 2cos2 dθ θ= −∫2 sin2 cθ θ= − +

We now need to substitute theta in terms of x.

2 2sin cos cθ θ θ= − +

2sinx θ=

sin2xθ = ÷

1sin2xθ − = ÷

2

22 2sin cos

4

xdx c

xθ θ θ= − +

−∫

2 24sinx θ=24 4cos θ= −

2 24cos 4 xθ = −22cos 4 xθ = −

21cos 4

2xθ = −

1 212sin 2. . 4

2 2 2x x

x c− = − − + ÷

1 22sin 42 2x x

x c− = − − + ÷

Now for some not very obvious substitutions……………….

51. Find sin x dx∫5 4sin sin sinx x x= 2 2sin (sin )x x= 2 2sin (1 cos )x x= −

5 2 2sin sin (1 cos )x dx x x dx= −∫ ∫ Let cosu x= sindu x dx= −

5 2 2sin (1 )x dx u du= − −∫ ∫ 2 4(1 2 )u u du= − − +∫ 2 41 2u u du= − + −∫3 52 1

3 5u u u c= − − +

3 52 1cos cos cos

3 5x x x c= − − +

12. Find

1dx

x−∫ Let 1u x= −

121

2du x dx

−= −

122dx x du⇒ = − ( )2 1dx u du⇒ = − −

1 2( 1)

1

udx du

ux

−=−∫ ∫ 2

2 duu

= −∫2 2lnu u c= − +

2 2 2ln 1x x c= − − − +

Integration by PartsIntegration by Parts

Integration By Parts

Start with the product rule:

( )d dv duuv u v

dx dx dx= +

( ) d uv u dv v du= +

( ) d uv v du u dv− =

( ) u dv d uv v du= −

( )( ) u dv d uv v du= −∫ ∫

( )( ) u dv d uv v du= −∫ ∫ ∫

This is the Integration by Parts formula.

→

∫ ∫ ′−=′ dxvuuvdxvu

The Integration by Parts formula is a “product rule” for integration.

u differentiates to zero (usually).

v’ is easy to integrate.

Choose u in this order: LIPET

Logs, Inverse trig, Polynomial, Exponential, Trig

→

∫ ∫ ′−=′ dxvuuvdxvu

Example 1:

polynomial factor u x=sinv x=

LIPET

sin sin x x x dx⋅ − ∫

→

∫ dxxx cos

xv cos=′

∫ ∫ ′−=′ dxvuuvdxvu

1=′u∫ ∫ ′−=′ dxvuuvdxvu

cxxxdxxx ++=∫ cossincos

Example 2:

logarithmic factor

LIPET

→

∫ dxxln

1=′v

∫ ∫ ′−=′ dxvuuvdxvu

xu

1=′∫ ∫ ′−=′ dxvuuvdxvu

cxxxdxx +−=⋅∫ ln1ln

xu ln=

xv =

1ln x x x dx

x⋅ − ⋅∫

This is still a product, so we need to use integration by parts again.

Example 3:

2 xx e dx∫LIPET

2u x=xv e=

u x=xv e=

→

∫ ∫ ′−=′ dxvuuvdxvu

xu 2=′

xev =′

1=′u

xev =′

∫ ′−= dxvuuv

∫−= dxxeex xx 22

∫−= dxxeex xx 22

( )∫−−= dxexeex xxx 22

cexeexdxex xxxx ++−=∫ 2222

∫ ∫ ′−=′ dxvuuvdxvu

u dv uv v du= −∫ ∫

The Integration by Parts formula can be written as:

Example 4:

cos xe x dx∫LIPET

xu e= sin dv x dx= xdu e dx= cosv x= −

u v v du− ∫sin sinx xe x x e dx− ×∫

( )sin cos cos x x xe x e x x e dx− ×− − − ×∫

xu e= cos dv x dx= xdu e dx= sinv x=

sin cos cos x x xe x e x e x dx+ − ∫This is the expression we started with!

→

uv v du

Example 4 continued …

cos xe x dx∫LIPET

u v v du− ∫

cos xe x dx =∫2 cos sin cosx x xe x dx e x e x= +∫

sin coscos

2

x xx e x e xe x dx C

+= +∫

sin sinx xe x x e dx− ×∫xu e= sin dv x dx=

xdu e dx= cosv x= −

xu e= cos dv x dx= xdu e dx= sinv x=

sin cos cos x x xe x e x e x dx+ − ∫

( )sin cos cos x x xe x e x x e dx− ×− − − ×∫

Example 4 continued …

cos xe x dx∫ u v v du− ∫

This is called “solving for the unknown integral.”

It works when both factors integrate and differentiate forever.

→

cos xe x dx =∫2 cos sin cosx x xe x dx e x e x= +∫

sin coscos

2

x xx e x e xe x dx C

+= +∫

sin sinx xe x x e dx− ×∫

sin cos cos x x xe x e x e x dx+ − ∫

( )sin cos cos x x xe x e x x e dx− ×− − − ×∫

A Shortcut: Tabular Integration

Tabular integration works for integrals of the form:

( ) ( )f x g x dx∫where: Differentiates to zero

in several steps.Integrates repeatedly.

→

2 xx e dx∫( ) & deriv.f x ( ) & integralsg x

2x

2x

2

0

xexexexe

+

+

−

2 xx e dx =∫ 2 xx e 2 xxe− 2 xe+ C+

Compare this with the same problem done the other way:

→

This is still a product, so we need to use integration by parts again.

Example 3:

2 xx e dx∫LIPET

2u x=xv e=

u x=xv e=

→

∫ ∫ ′−=′ dxvuuvdxvu

xu 2=′

xev =′

1=′u

xev =′

∫ ′−= dxvuuv

∫−= dxxeex xx 22

∫−= dxxeex xx 22

( )∫−−= dxexeex xxx 22

cexeexdxex xxxx ++−=∫ 2222

This is easier and quicker to do with tabular integration!

3 sin x x dx∫3x

23x

6x

6

sin x

cos x−

sin x−cos x

+

+

−

0

−

sin x

3 cosx x− 2 3 sinx x+ 6 cosx x+ 6sin x− + C

π

Properties of the Definite Integral

1:

2:

3:

4:

5:

6:

7:

8:

9:

10:

11:

12:

Substitution and definite integrals Assuming the function is continuous over the interval, then exchanging the limits for x by the corresponding limits for u will save you having to substitute back after the integration process.

22 3

1

1. Evaluate (2x+4)(x 4 )x dx+∫2Let 4u x x= + 2 4du x dx= + When 2, 12; 1, 5x u x u= = = =

2 122 3 3

1 5(2x+4)(x 4 )x dx u du+ =∫ ∫

124

5

14u =

5027.75=

Special (common) formsSome substitutions are so common that they can be treated as standards and, when their form is established, their integrals can be written down without further ado.

1( ) ( )f ax b dx F ax b c

a+ = + +∫`( )

ln ( )( )f x

dx f x cf x

= +∫21

`( ) ( ) ( ( ))2

f x f x dx f x c= +∫

Area under a curve

a b

y = f(x)

( )b

aA f x dx= ∫

a b

y = f(x)

( )b

aA f x dx= −∫

Area between the curve and y - axis

b

a

y = f(x)

( )b

aA f y dy= ∫

1. Calculate the area shown in the diagram below.

5

2

y = x2 + 1

2 1y x= +2 1x y= −

1x y= −

15 2

2( 1)A y dy= −∫

532

2

2( 1)

3y

= −

14 units squared.

3=