Transcript Applications Of The Definite Integral

Applications 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 Curve
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 :
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 xaxis and the lines x=a and x=b for each of the
following cases:
a=-2 b=2
a=2
b=3
2
a=-2 b=3
2
3
-2
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]
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)
2
A1

f ( x) d x
2
2


2  x dx
A1
2
2
2

 (2 
x)d x
-2
2
2
x2
 (2 x 
)
2
2
4
4
 (4 
)  ( 4 
)
2
2
 2  6  8
2
3
Case2
The area A2 between f, the x-axis and the
Lines x=2 and x=3 is:
3
f(x)<0; x (2, 3]
A1  f ( x) dx
2
3
  2  x dx
2
2
3
x2 3
   (2  x)dx  (2 x  ) 2 dx
2
2
9
4
 (6  )  (4  )
2
2
 1/ 2
-2
2
3
Case3:
The area a between f, the X-axis and the lines
X=-2 and X=3 is :
3

f ( x) d x
2
2
3


2  x dx
2
2

-2
 (2  x)d x  
2
 8 1/ 2
 17 / 2
3
2
 (2  x)d x
2
3
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)
5
Let f (x) =X, g (X) = 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
5
x
So |g (X) –f (X)| =g (X)-f (X)= -x
Case (1):
The area A1 between f and g from X= -1 and x=0
g
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 5 -x
1
0
A1  g ( x)  f ( x) dx
1
1
1
0
  ( x 5  x)dx
1
 ( x / 6  x / 2)
6
1
2
0
1
 (0  0)  (1 / 6  1 / 2)
 1/ 3
Case (2)
g
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
5
F(X) –g (X)>0 so
|g (X) –f (X)| =f (X) –g (X) =x- x
1
1
A2   g ( X )  f ( X ) dx
0
 x  x dx
5
0

1
1
1

1
 x2 / 2  x6 / 6

1
0
1 1
1
 (  )  0  0  
2 6
3
f
Case (3)
g
The area A between f and g from X = -1 to X=1
1
A3  g ( x)  f ( x) dx
1
1
0
1
  ( x  x )   ( x  x )dx
5
1
 1/ 3 1/ 3
 2/3
0
5
1
1
1
f
Volumes of Revolution :
2
f
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.
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 
x2
  f
2
( x) d x
x1

2

x
dx
0
2
2


x2d x
0


 [ (x 3
8 / 3
2
/ 3]0
Example1:1
Consider the area bounded by the graph of the function
2
x
f(x) =x- and x-axis:
1
The volume of solid is:
1
2 2

(
x

x
) dx

0
1
   ( x 2  2 x 3  x 4 )dx
0
  ( x / 3  2 x  x / 5)
3
3
5
1
0
  (1 / 3  2 / 4  1 / 5)   (0 / 3  0 / 4  0 / 5)
  / 30
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