DEFINITE INTEGRALS

OBJECTIVES

At the end of the lesson, the students are expected to:• define and interpret definite integral.• identify and distinguish the different properties of the definite

integrals. • evaluate definite integrals.• understand and use the mean Value Theorem for Integrals.• find the average value of a function over a closed interval.

If F(x) is the integral of f(x)dx, that is, F’(x) = f(x)dx and if a and b are constants, then the definite integral is:

)a(F)b(F

xF dx)x(f ba

b

a

where a and b are called lower and upper limits of integration, respectively.

The definite integral link the concept of area to other important concepts such as length, volume, density, probability, and other work.

THE DEFINITE INTEGRAL

0 1

23 2 xy

It can be used to find an area bounded, in part, by a curvee.g.

1

0

2 23 dxx gives the area shaded on the graph

The limits of integration . . .

Definite integration results in a value.

Areas

. . . give the boundaries of the area.

The limits of integration . . .

0 1

23 2 xy

It can be used to find an area bounded, in part, by a curve

Definite integration results in a value.

Areas

x = 0 is the lower limit( the left hand boundary )

x = 1 is the upper limit(the right hand boundary )

dx x 23 2

0

1e.g.

gives the area shaded on the graph

0 1

23 2 += xy

Finding an area

the shaded area equals 3

The units are usually unknown in this type of question

1

0

2 23 dxxSince

31

0

xx 23

xxy 22 xxy 22

Finding an area

0

1

2 2 dxxxA area

A B

1

0

2 2 dxxxB area

For parts of the curve below the x-axis, the definite integral is negative, so

EXAMPLE

•

•

EXERCISES

INTEGRATION OF ABSOLUTE VALUE FUNCTION

0 xif

0 xif x Re

x

xcall

INTEGRATION OF ABSOLUTE VALUE FUNCTION

EXAMPLE dx x .14

2

1082

02

)4(

2

)2(0

22

x

224

0

20

2

2

4

0

0

2

4

2

xx

xdxxdxdx

1st solution

0 xif

0 xif x

x

x

0 1-2 32 4-1

0x0x

2nd solution

(4,4) 4; (4) f

(0,0) 0; (0) f

,2) (-2 2;f(-2)

y)(x,

xf(x)

let

-1 1-2 432

(4,4)

(-2,2)

0

8)4)(4(2

1

2)2)(2(2

1

2

1

A

A

10

82

21

4

2

AAdxx

3

31 .2 dxx

1st solution

1 x0,x-1 if 1

1 x0,x-1 if 11

x

xx

102

20

2

1

2

3

2

15

2

1

22

x-x

111

3

1

21

3

2

3

1

1

3

3

3

xx

dxxdxxdxx

2nd solution

(3,2) 2; (3) f

(1,0) 0; (1) f

,4) (-3 4;f(-3)

y)(x,

x-1f(x)

let

-1 1-2-3 32

(3,2)

(-3,4)

0

2)2)(2(2

1

8)4)(4(2

1

2

1

A

A

10

28

1 21

3

3

AAdxx

EXERCISES

INTEGRATION OF ODD AND EVEN FUNCTIONS

x- integers, oddFor

;x- integers,even For

:Re

nn

nn

x

x

call

f ofdomain xallfor f(x)f(-x) ifeven be tosaid is Function

f ofdomain xallfor f(x)f(-x) if odd be tosaid is Function

The graph of an even function is symmetric about the y-axis.

The graph of an odd function is symmetric about the origin.

oddisfunctionthebecausedtt

t0

1

3

3 2

3

EXAMPLE

INTEGRATION OF PIECEWISE FUNCTION

EXAMPLE

4x0 ,2

2

1

0x2- ,2 f(x) ; )( .1

2

4

2 x

xdxxf

solution

3

56x2

4

x

3

xx2

dx2x2

1dxx2dx)x(f

4

0

20

2

3

0

2

4

0

24

2

6

6

0

0

2-

.

4

2

2)(.)(.

)()(.

0,2

0,2)(

int,.1

dxxfddxxfb

dxxfcdxxfa

xifx

xifxxf

thatgivenegraltheevaluateparteachIn

EXERCISES

5

10

1

1

0

.

2

1

1

1)(.)(.

)()(.

1,2

1,2)(

int,.2

dxxfddxxfb

dxxfcdxxfa

xif

xifxxf

thatgivenegraltheevaluateparteachIn

EXERCISES

b c

Area =0.8

Area =2.6

Area =1.5

d

a

c

b

c

a

b

a

dx)x(f .d dx)x(f .b

f(x)dx .c dx)x(f .a

find to figure the in shown areas the Use

da

Answers :

0.8 dx)x(f .ab

a

2.6 dx)x(f .bc

b

-0.32.6-1.50.8 dx)x(f .d d

a

1.8- 2.6-0.8 f(x)dx .cc

a

OTHER EXAMPLE

Find the definite integral of the following

The mean value theorem for integrals state that somewhere “between” the inscribed and the circumscribed rectangles there is a rectangle whose area is precisely equal to the area of the region under the curve.

EXRCISES

Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval.

Find the average value of the function over the given interval.

EXRCISES