IntegrationIntegration

Indefinite Integral

Suppose we know that a graph has gradient –2, what is the equation of the graph?

There are many possible equations for graphs with gradient –2:

2y x

2 1y x

2 1y x

2 2y x

Indefinite Integral

Suppose now we have

What are the possible equations for y what will give ?

4dy

xdx

22 2y x

22y x

22 1y x

The process of obtaining the equation of the curve, given its derivative, is called Integration (or the antiderivative).

Hence, Integration is the reverse process of differentiation.

( )dy x dx y

dxx

Consider:2 3y x

then: 2y x 2y x

2 5y x or

It doesn’t matter whether the constant was 3 or -5, since when we take the derivative the constant disappears.

However, when we try to reverse the operation:

Given: 2y x find y

2y x C

We don’t know what the constant is, so we put “C” in the answer to remind us that there might have been a constant.

Indefinite Integral

1, and are constants,n a n

naxdx

dy

1

1

n

n

y ax dx

xan

c

In general, if

,where c is a constant.

Example 1: Integrate the following expressions with respect to x

2( )a x ( )b x ( ) 4c

1

1

n

n

y ax dx

xan

c

2x dx3

3

xc

,where c is a constant.

1

2x dx3

2

3

2

3 / 2

2

3

xc

xc

,where c is a constant.

4 dx4x c

,where c is a constant.

cn

axdxax

nn

1

1

Indefinite Integral :

, where c is an arbitrary constant and n 1.

Some useful rules:

0 dx C1dx x C a dx ax C

Example 2: Find y in terms of x, of the following:

2( ) 27dy

a xdx

3( ) 8dy

b xdx

2

5( )dy

cdx x

2

3

3

27

27

3

9

x dx

xc

x c

3

1

3

4

3

4

3

8

8

8

4 / 3

6

x dx

x dx

xc

x c

2

2

1

5

5

5

15

dxx

x dx

xc

cx

,where c is a constant.

,where c is a constant. ,where c is a constant.

If we have some more information we can find C.

Given: and when , find the equation for .2y x y4y 1x

2y x C 24 1 C

3 C2 3y x

This is called an initial value problem. We need the initial values to find the constant.

An equation containing a derivative is called a differential equation. It becomes an initial value problem when you are given the initial condition and asked to find the original equation.

Integration of ( )nax b38(2 3)x 4(2 3)

dx

dx

34(2 3 2)) (x

Multiply by du

dxOutside first!

48(2 3

2)

)

(4

x

Outside first!

Divide by du

dx

Chain Rule: Apply when you have a composite function: f[g(x)]*n > 1 and coefficient of x ≠ 1

cna

baxc

baxdx

dn

baxbax

nnn

)1()().1(

11

, where c is a constant.

Example 8: Integrate the following expressions with respect to x

4( ) (5 1)a x 3( ) (1 )b x 2

1( )

2 1c

x

( ) 1 4d x

4(5 1)x dx5(5 1)

(5)(5)

xc

5(5 1)

25

xc

3(1 )x dx4(1 )

(4)( 1)

xc

4(1 )

4

xc

,where c is a constant.

,where c is a constant.

2(2 1)x dx1(2 1)

( 1)(2)

1

2(2 1)

xc

cx

,where c is a constant.

1

2(1 4 )x dx3

2

3

2

(1 4 )3

( 4)2

(1 4 )

6

xc

xc

,where c is a constant.

3.2 Definite Integrals

b

af x dx

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

[ ( )]

( ) ( )

b baa

f x dx F x

F b F a

434 2

33

43

3

4 43 3

3 3

33

3

64 27

4

3

3

7

xx dx c

x c

c

c c

c

So far when integrating, there has always been a constant term left. Hence, such integrals are known as indefinite integrals. With definite integrals, we integrate a function between 2 points, and so we can find the precise value of the integral and there is no need for any unknown constant terms [the constant cancels out].

Definite Integrals

Example:

Example 9: Evaluate

2

1

( ) 6 2a x dx4

0( ) 2 1b x dx

2

1

( ) ( 1)( 2)c x x dx

22

13 2x x

(12 4) (3 2)

16 5

11

4 1

2

0

4

3

2

0

(2 1)

2 1

3(2)

2

x dx

x

43

2

0

2 1

3

x

19

3

28

3

Expect a numerical answer!

22

1

23 2

1

( 3 2)

32

3 2

x x dx

x xx

38 23

3 653

6

1

2 33

1 1( )d dx

x x

12 3

3

11 2

3

1

23

( )

1 2

1 1

2

x x dx

x x

x x

3 7

2 1810

9

Example 10:

If 0

( 4)ax dx = 0, find the values of a.

2

00

2

( 4) 42

4 102

aa xx dx x

aa

2 8 20 0

( 10)( 2) 0

10 2

a a

a a

a or

Example 11: Given that 1)2( xxy .Show that 12

3

x

x

dx

dy

Hence evaluate 8

3 1

xdx

x

1)2( xxy

Show :dy

dx

1 1

2 2

1

2

1( 2) ( 1) ( 1)

2

1( 1) ( 1 1)

23

2 1

dyx x x

dx

x x x

x

Must make use of 1st part result!

From 1st part, we know:

3( 2) 1

2 1

dy xy x x

dx x

3( 2) 1

2 1

xdx x x c

x

Hence,

88

33

3( 2) 1

2 1

xdx x x

x

88

33

2( 2) 1

31

26 9 4

332

3

xdx x x

x

Example 12: Given that .Show that

Hence evaluate

x

xy

1

23 2

2

2

)1(

263

x

xx

dx

dy

2

1 2

2

12

363dx

x

xx

23 2

1

xy

x

Show :dy

dx

2

2

2 2

2

2

2

(1 )6 (3 2)( 1)

(1 )

6 6 3 2

(1 )

3 6 2

(1 )

dy x x x

dx x

x x x

x

x x

x

From 1st part, we know:

2 2

2

3 2 3 6 2

1 (1 )

x dy x xy

x dx x

Hence,

2

1 2

2

12

363dx

x

xx

Hence,

2

1 2

2

12

363dx

x

xx

22

21

22

21 2

3 6

1

3 6

1 2 1

2

1 2 1

2 11 x

x xdx

x

x xdx

x

2 2

211

2 21

1 1

2 2

2

2

11

2

1 1 1

2 2 1

1 1

3 2

1

(1 )

2 2 ( 1)( 1)

1 1 1

2 2 1

3 2

1

3 2

1

9

dxx

x

x

x

x

x

x

x

x

2. 0a

af x dx If the upper and lower limits are equal,

then the integral is zero.

1. b a

a bf x dx f x dx Reversing the limits

changes the sign.

b b

a ak f x dx k f x dx 3. Constant multiples can be

moved outside.

Useful Rules:

b b b

a a af x g x dx f x dx g x dx 4.

Integrals can be added and subtracted.

Using ,)()()( b

a

aFbFdxxf

5. b c c

a b af x dx f x dx f x dx

Intervals can be added(or subtracted.)

a b c

y f x

211

8y x

43

0

1

24A x x

4 2

0

11

8A x dx

0 4x

Actual area under curve:

20

3A 6.6

Example 2:

Area under a curve – Estimate using Trapezoid Rule

Trapezoid RuleDivide curve into series of trapezoids

1 1 1 1 1 11 (0 6) 1 (6 16) 1 (16 30) 1 (30 48) 1 (48 70) 1 (70 96)

2 2 2 2 2 2218

Area

0

20

40

60

80

100

120

1 2 3 4 5 6 7

(1,0)(2,6)

(3,16)(4,30)

(5,48)

(6,70)

(7,96)

Area under a curve – Estimate using Simpson’s Rule

0 4(6) 16 40

21

1 231,3,5,

Simpson's 1/3 Rule

( 4 )n

i i ii

I y y y x

Simpson's 1/3 RuleSeven points making three strips

0

20

40

60

80

100

120

0 1 2 3 4 5 6 7 8

(1,0)

(3,16)

(5,48)

(7,96)

(2,6)

(4,30)

(6,70)

Area under Graphs – straight line

x

y

y = f(x) = x

x0

Area under the line y = x can be found using area of triangle

Area =

2

1

21

( )( )2

2

base height

x x

x

0

2

0

2

2

2

x

x

x dx

x

x

Let’s use another method to find the area under the line:

x

yy = f(x) = x

10

Area =

We will use rectangles to approximate the area. Let’s start with n rectangles

Hence, width of each

rectangle = 1

n1

n

1

n

1

n

1

n

3.3 Integration by Substitution

52x dx Let 2u x

du dx5u du61

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!

Example 1:

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 dx

1

2 u du3

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.

Example 2:

4 1 x dx Let 4 1u x

4 du dx

1

4du dx

1

21

4

u du3

22 1

3 4u C

3

21

6u C

3

21

4 16

x C

Example 3:

=

=

=

1 2 3

13 x 1 x dx

3Let 1u x

23 du x dx1

22

0 u du

23

2

0

2

3u

Don’t forget to use the new limits.

3

22

23

22 2

3 4 2

3

Example 4:

=

=

when x = -1, u = 0

when x = 1, u = 2

21 2y x

2y x

How can we find the area between these two curves?

We could split the area into several sections, use subtraction and figure it out, but there is an easier way.

Area under the curves

21 2y x

2y x

Consider a very thin vertical strip.

The length of the strip is:

1 2y y or 22 x x

Since the width of the strip is a very small change in x, we

could call it dx.

21 2y x

2y x

1y

2y

1 2y ydx

Since the strip is a long thin rectangle, the area of the strip is:

2length width 2 x x dx

If we add all the strips, we get:2 2

12 x x dx

21 2y x

2y x

2 2

12 x x dx

23 2

1

1 12

3 2x x x

8 1 14 2 2

3 3 2

8 1 16 2

3 3 2

36 16 12 2 3

6

27

6

9

2

The formula for the area between curves is:

1 2Areab

af x f x dx

We will use this so much, that you won’t need to “memorize” the formula!

y x

2y x

y x

2y x

If we try vertical strips, we have to integrate in two parts:

dx

dx

2 4

0 2 2x dx x x dx

We can find the same area using a horizontal strip.

dySince the width of the strip is dy, we find the length of

the strip by solving for x in

terms of y.y x2y x

2y x

2y x

y x

2y x

We can find the same area using a horizontal strip.

dySince the width of the strip is dy, we find the length of

the strip by solving for x in

terms of y.y x2y x

2y x

2y x

2 2

02 y y dy

length of strip

width of strip

22 3

0

1 12

2 3y y y

82 4

3

10

3

General Strategy for Area Between Curves:

1

Decide on vertical or horizontal strips. (Pick whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.)

Sketch the curves.

2

3 Write an expression for the area of the strip.(If the width is dx, the length must be in terms of x. If the width is dy, the length must be in terms of y.

4 Find the limits of integration. (If using dx, the limits

are x values; if using dy, the limits are y values.)

5 Integrate to find area.