Application Of Definite Integrals

APPLICATIONS OF THE DEFINITE INTEGRAL The definite integral of a function are used in many applications. Those discussed here are •

Areas bounded by curves,

•

Volumes

•

Lengths of arcs of curves.

AREAS The area of a region bounded by the graph of a function, the x-axis, and two vertical lines can be done by using definite integral. If in an interval [a,b], f x ≥ 0 , then the region ` a

` a

lying below the graph of f x , above the x-axis and between the two vertical lines x = a and x = b is given by b

A = Z f x dx a

` a

Figure: Finding the area under a positive function

If f x ≤ 0 on [a,b], then the area of the region lying above the graph of f x , below ` a

` a

the x-axis and between the two vertical lines x = a and x = b is given by b

A = @ Z f x dx = a

` a

L M L b M L M L ` a M L Z f x dxM L M L M La M

Figure : Finding the area above a negative function

Note that in both the cases the graph does not intersect the x-axis. If the graph of f x intersects x-axis at x = c, where a < c < b , and if f x ≥ 0 on [a,c] ` a

` a

and f x ≤ 0 on [c,b] , then the area of the region bounded by the graph of f x , the x` a

axis and the two vertical lines x = a and x = b is given by b

A =

L ` aM M ZL L f x Mdx a

c

b

= Z f x dx @ Z f x dx a

` a

c

` a

` a

Figure : The area bounded by a function whose sign changes.

In such situations it is necessary to determine all the points where the function changes sign , that is , the points where the curve intersects the x-axis, and to determine the sign of f(x) in each interval.

We now move on to find areas of regions bounded by the graphs of two or more functions, or two functions and two vertical lines x = a and x = b . First, we need to locate the position of each graph relative to the position of the other. The points of intersection of the graphs might be necessary.

For example, if f x ≥ g x ` a

` a

at all points on [a,b] , then the area between the graphs of

f(x) and g(x) and two vertical lines x = a and x = b is given by b

A = Z f x @ g x dx a

B ` a

` aC

Figure : The area between two functions.

Area with the y-axis : We can also find the area between graph of horizontal lines y = a and y = b.

x = f y , the y-axis and between the ` a

If in an interval [a,b], f y ≥ 0 , then the region lying on the left of the graph of f y , ` a

` a

on the right of the y-axis and between the two horizontal lines y = a and y = b is given by b

A = Z f y dy a

` a

Similarly, in an interval [a,b], f y ≤ 0 , then the region lying on the right of the graph ` a

` a

of f y , on the left of the y-axis and between the two horizontal lines y = a and y = b is given by b

A = @ Z f y dy a

` a

Example : Find the area of the region bounded by y = x 2 , the x @ axis, x = @ 3 and x = 2 A

Solution : Here f x = x 2 and f x ≥ 0 in the interval @ 3, 2 ` a

So, the area

2

` a

A = Z x 2 dx @3

B C2 1f f3 = f x 3 @3 B a3C 1f f 3 ` = f 2 @ @3 3 35 f f f f f f = f 3

B

C

Example : Find the area of the region bounded by the curve y = x 3 + x 2 @ 6x and the xaxis.

Solution : Below is the graph of the function.

Here y = f x = x 3 + x 2 @ 2x To know where the curve intersects the x @ axis, we put y = 0 in the equation, x 3 + x 2 @ 2x = 0, ` a

b

c

x x 2 + x @ 2 = 0, x x + 2 x @1 =0 x = 0, x = @ 2, x = 1 `

a`

a

B

C

B

C

So the curve bounds an area with the x @ axis in the interval @ 2,0 and 0,1 A B

In the interval @ 2,0

f x ≥ 0 , and in the interval 0,1

C

B

` a

C

1

A = Z x 3 + x 2 @ 2x dx @2

B

C

0

1

= Z x + x @ 2x dx @ Z x 3 + x 2 @ 2x dx @2

B

3

2

C

0 0

=

F 1f f f f

=

8f f f F

=

37 1f f f f f f f f f f f f f f = 3 12 12

G 1f f f x + x3@ x2 4 3

3

4

@ @

@2

G 5f f f f f f f

12

@

B

C

F 1f f f f

1

G 1f f f x + x3@ x2 4 3 4

0

f x ≤0 ` a

Example : Find the area bounded by the curves y = x 2 @ 6 and y = 12 @ x 2 .

Solution :

To find the points of intersection of the two curves y = x 2 @ 6 and y = 12 @ x 2 , we equate the y, x 2 @ 6 = 12 @ x 2 2x 2 = 18 x 2 @ 9a=` 0 a ` x + 3 x @3 =0 x = 3, x = @ 3 at x = @ 3, y = 3 and at x = 3, y = 3 A So the curves intersect at @ 3,3 and 3,3 A As

12 @ x 2 ≥ x 2 @ 6 in the interval ` a

B

C

@ 3,3

Therefore the Area A of the region bounded by them 3

A

=

Z @3 3

Db

c

2

b

cE

2

12 @ x @ x @ 6 dx

b

c

= Z 18 @ 2x 2 dx @3

3

= = # A = 72

2f f 3G 18x @ f x 3

F

@3

54 @ 18 @ @ 54 + 18

@

A

@

A

= 72

b

c

b

c

VOLUMES OF SOLIDS WITH KNOWN CROSS SECTIONS If we know a formula for the region of the cross section perpendicular to an axis of a solid, we can use definite integral to find the volume of the solid in an interval on that axis. If the cross sections are perpendicular to the x-axis , then their areas will be a function of x. If we denote this function as A(x) then the Volume ( V ) of the solid on the interval [a,b] is given by b

V = Z A x dx a

` a

Similarly, if the cross sections are perpendicular to the y-axis , then their areas will be functions of y, which we denote as A(y) . Then the Volume ( V ) of the solid on the interval [a,b] is given by b

V = Z A y dy a

` a

Example : Find the volume of the solid whose base is the region bounded by the lines x + 5y = 0, x = 0 and y = 0, if cross sections taken perpendicular to the x-axis are squares.

Solutions :

` a

The Area A of an arbitrary square cross section is g2

`f a2 5f @ xf 1f f f f f f f f f f f f f f f f f f A x = f = f 5@x 5 25 f

` a

` a

The Volume V 5

of the solid is

V = Z A x dx ` a

0 5

`f a2 1f f f f f 5 @ x dx = Z f 25 0

5

b c 1f f f f f fZ 25 @ 10x + x 2 dx = f 25 0

5

3G xf f f f f f = 25x @ 5x + f 25 3

F 1f f f f f f f

2

0

F G 1f 125 5f f f f f f f f f f f f f f f f f = f 125 @ 125 + f = f 25 3 3

A = s2

5f @ xf f f f f f f f f f f f f f where s = f 5

ARC LENGTH By using definite integral, we can find the length of an arc along a curve. In this closed interval the function and its derivative both needs to be continuous.

If y = f x and y. = f. x are both continuous in the interval a,b ` a

` a

` a

then the Arc Length L

B

C

C

of f x on a,b is

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w B ` aC2 s = Z 1 + f. x dx b

L

` a

B

a

Similarly, If x = f y and x . = f. y are both continuous in the interval a,b then the ` a

` a

` a

` a

B

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w B ` aC2 s = Z 1 + f. y dy b

C

arc length L of f y on a,b is

L

B

Example : Find the arc length of the graph of

C

a

B C 3 ` a 1f f 2fffff f x = f x on the interval 0,5 3

3 ` a 1f f 2fffff f x = f x in which the required arc is highlighted. 3

Solution : Below is the graph of

f x = ` a

L

3 1f f f 2fffff x 3

v w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w 2 f g 1 f f f f f u 1f f f Z t1 + x 2 dx 5u u

=

2

1 ` a 1f f f fffff f. x = x 2 2 5

=

0

H

f 2f f f

=L J4 A 3 A 1 + L

=

F 8f f f27 f f f f f f f G @1 3 8

I5 3f f f f f g2 1f f f f M x M K 4

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w 1f f f f

Z s1

+

4

x dx

0

=

0

=

H f 2f f fL L 4 A AJ 1 3

3f f f f f

+

19 f f f f f f f 3

Example : Find the arc length of the graph of

g2 5f f f f

4

I M

@ 1M K

πf f f f f x = ln sec x on the interval 0, f 6 ` a

`

a

D

E

f x = ln sec x ` a sec xf tan xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f. x = f = tanx sec x ` a

L Solution :

`

a

πf f f f f 6 w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w ` a2

= Z q1 + tan x

dx

0 πf f f f f 6

= Z sec x dx 0

f

= ln

g ` a 2f 1f f f f f f f f f f f f f f f f f f f f f w w w w w w+ w w w w w w @ ln 1 p3 p3

w w w w w w 1f f = ln p3 = f ln 3 2 = 0.55

πf f f f f 6 w w w w w w w w w w w w w w w w w w w w w w w w w

= Z qsec 2 x dx 0

= ln sec x + tan x B `

πf f f f f aC6 0