Applications Of The Definite IntegralApplications Of The Definite Integral

The Area under the curve of a function

The area between two curves

The Volume of the Solid of revolution

Applications of the Definite Integral

• In calculus, the integral of a function is an extension of the concept of a sum. The process of finding integrals is called integration. The process is usually used to find a measure of totality such as area, volume, mass, displacement, etc.

The integral would be written . The∫ sign represents integration, a and b are the endpoints of the interval, f(x) is the function we are integrating known as the integrand, and dx is a notation for the variable of integration. Integrals discussed in this project are termed definite integrals.

We use definite integrals

• basically,

• Results:

               

Given:  

evaluate,

Solution:

Area under a CurveTo find the area under a curve. This expression gives us a

definite value (a number) at the end of the calculation.

When the curve is above the ‘x’ axis, the area is the same as the definite integral :

But when the graph line is below the ‘x’ axis, the definite integral is negative. The area is then given by:

(Positive)

(Negative)

Example 1:

let f (x)=2-x .Find the area bounded by the curve of f , the x-axis and the lines x=a and x=b for each of the following cases:

a=-2 b=2

a=2 b=3

a=-2 b=3

The graph:

Is a straight line y=2-x:

F (x) is positive on the interval [-2, 2)

F (x) is negative on the interval (2, 3]

2

2 3-2

Case 1:

The area A1 between f, the x-axis and the

lines x=-2 and x=2 is: f(x)>0; x [-2,2)

862

)2

44()

2

44(

)2

2(

)2(

2

)(

2

2

2

2

2

2

2

2

2

1

xx

dxx

dxx

dxxfA

2

32-2

A1

f(x)<0; x (2, 3]

Case2

The area A2 between f, the x-axis and the

Lines x=2 and x=3 is:

2/1

)2

44()

2

96(

)2

2()2(

2

)(

32

23

2

3

2

3

2

1

dxx

xdxx

dxx

dxxfA

3

2

2-2

Case3: The area a between f, the X-axis and the lines

X=-2 and X=3 is :

2/17

2/18

)2()2(

2

)(

3

2

2

2

3

2

3

2

dxxdxx

dxx

dxxf

2

2

3-2

Area Bounded by 2 Curves

Say you have 2 curves y = f(x) and y = g(x)

Area under f(x) =

                              

Area under g(x) =

                              

Superimposing the two graphs, Area bounded by f(x) and g(x)

Example (2)

Let f (x) =X, g (X) = Find the area between f and g from X=a to X=b Following cases

a=-1 b=0

a=0 b=1

a=-1 b=1

g (X)>f (X) on (-1,0) and hence on this interval

, we have: g (X) –f (X)>0

So |g (X) –f (X)| =g (X)-f (X)= -x 5x

5x

Case (1):The area A1 between f and g from X= -1 and x=0

is: is:g (X)>f (X) on (-1,0) and hence on this interval

, we have :

g (X) –f (X)>0

So |g (X) –f (X)| =g (X)-f (X)= -x

3/1

)2/16/1()00(

)2/6/(

)(

)()(

0

1

26

50

1

0

1

1

xx

dxxx

dxxfxgA1

1

11

g

5x

Case (2)

The area A between f and g from X = 0 to X=1f(x) >g (X) on(0,1) and hence on this interval

, we have

F(X) –g (X)>0 so |g (X) –f (X)| =f (X) –g (X) =x-

3

100)

6

1

2

1(

6/2/

)()(

1

0

62

1

0

5

1

0

2

xx

dxxx

dxXfXgA

1

1

11

gf

5x

Case (3)

The area A between f and g from X = -1 to X=1

3/2

3/13/1

)()(

)()(

1

0

550

1

1

1

3

dxxxxx

dxxfxgA

1

1

11

gf

Volumes of Revolution :V=Π∫ (x)dx

• A solid of revolution is formed when a region bounded by part of a curve is rotated about a straight line.

2f

Rotation about x-axis:

Rotation about y-axis:

Example: Find the volume of the solid generated by revolving the region bounded by the graph of

y = x, y = 0, x = 0 and x = 2. At the solid

Solution: we shall now use definite integrals to find the volume defined above. If we let

f(x) = x according to 1 above, the volume is given by the definite integral

Volume

3/8

]3/[(

)(

20

3

2

0

2

2

0

2

22

1

x

dxx

dxx

dxxfx

x

Example1:1

Consider the area bounded by the graph of the function

f(x) =x- and x-axis:

The volume of solid is:

30/

)5/04/03/0()5/14/23/1(

)5/23/(

)2(

)(

1

0

533

431

0

2

1

0

22

xxx

dxxxx

dxxx

1

2x

In conclusion, an integral is a mathematical object that can be interpreted as an area or a

generalization of area. Integrals, together with derivatives, are the fundamental objects of calculus. Other words for integral include

antiderivative and primitive.

The group members:

• Sabrina Kamal ___________________ID:2004/58527• Manal Alsaadi ____________________ID:2004/51562• Taiba Mustafa ____________________ID:2005/50524• Muneera Ahmed__________________ID:2004/550244

Math119 - Section (1)Fall 2006

Dr.F.K.Al-Muhannadi