Applications of Integrations

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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
 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 ∫ f(x) . 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.
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Area under a CurveArea under a Curve
[ ] )()()()( aFbFxFdxxf
b
a
b
a
−==∫
To 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 :
y= f(x)
Area = ∫
b
a
dxxf )(
x
Y
x = a x= b

But when the graph line is below the ‘x’ axis, the definite
integral is negative. The area is then given by:
y= f(x)
Area = ∫−
b
a
dxxf )(

(Positive)
(Negative)
2
1
0
2
1
2
1
1
0
1
0
2
=−=





=∫ xxdx
1−
1
11−
2
1
2
1
0
2
1
0
1
0
1
2
−=−=





=
−−
∫ xxdx

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
4
4()
2
4
4()
2
2(
)2(
2
)(
2
2
2
2
2
2
2
2
2
1
=+=
−−−−=−=
−=
−=
=
−
−
−
−
∫
∫
∫
x
x
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
4
4()
2
9
6(
)
2
2()2(
2
)(
3
2
23
2
3
2
3
2
1
=
+−−+−=
+−=−−=
−=
=
∫
∫
∫
dx
x
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
=
+=
−−+−=
−=
∫∫
∫
−
−
dxxdxx
dxx
2
2
3-2
∫−
3
2
)( dxxf

Area Bounded by 2 CurvesArea Bounded by 2 Curves
Area under f(x) =
Area under g(x) =
∫
b
a
dxxf )(
∫
b
a
dxxg )(
Say you have 2 curves y = f(x) and y = g(x)

Superimposing the two graphs, Area
bounded by f(x) and g(x)
∫
∫∫
−=
−
b
a
b
a
b
a
dxxgxf
dxxgdxxf
)()(
)()(

Example (2)
Let f (x) = x , g (x) = x5
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) = x5
- x

Case (1):
The area A1 between f and g from X= -1 and x=0 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) = x5
- x
3/1
)2/16/1()00(
)2/6/(
)(
)()(
0
1
26
0
1
5
0
1
1
=
−−−=
−=
−=
−=
−
−
−
∫
∫
xx
dxxx
dxxfxgA 1−
1
11−
g
f

Case (2)
The area A between f and g from x = 0 to x = 1
f(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 - x5
( )
( )
( )
3
1
00
6
1
2
1
6/2/
)()(
1
0
62
1
0
5
1
0
2
=−−





−=
−=
−=
−=
∫
∫
xx
dxxx
dxxfxgA
1−
1
11−
g
f

Case (3)
The area A between f and g from x = -1 to x =1
3/2
3/13/1
)()(
)()(
1
0
5
0
1
5
1
1
3
=
+=
−+−=
−=
∫∫
∫
−
−
dxxxxx
dxxfxgA
1−
1
11−
g
f

Volumes of Revolution :Volumes of Revolution :V=Π ∫fV=Π ∫f22
(x) dx(x) dx
 A solid of revolution is formed when a region bounded
by part of a curve is rotated about a straight line.
Rotation about x-axis:
Rotation about y-axis:
dxyV
b
a
∫Π= 2
dyxV
d
c
∫Π= 2

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:
Volume
[ ]
3/8
]3/[(
)(
2
0
3
2
0
2
2
0
2
2
2
1
π
ππ
ππ
=
==
==
∫
∫∫
xdxx
dxxdxxf
x
x
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

Example 1:1
Consider the area bounded by the graph of the function
f(x) = x – x2
and x-axis:
The volume of solid is:
30/
)5/04/03/0()5/14/23/1(
)5/23/(
)2(
1
0
533
43
1
0
2
π
ππ
π
π
=
+−−+−=
+−=
+−= ∫
xxx
dxxxx
1
∫ −
1
0
22
)( dxxxπ

• 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 anti-
derivative and primitive.
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