181
CHAPTER FIVE
Integration
and its Economic Applications
Conceptes of integration:
In the derivative we known the function Y f x and
we calculate the rate of change of function is represent the
dY
first derivative of this function or f X or Y .
dX
Integration is the reverse of differenatiation (
antiderivative). In many problems , the derivative of a
function is known and the goal is to find the function it self,
Def. a function F(x) for which f X f X for every
x in the domain of f is said to be an anti derivative ( or
indefinite integral) of F.
Integration Notation :
It is used to write: f X . dX f X C
- is called the integral sign.
- The function f(x) that is to be integrated is called the
integrand.
- C is called the constant of integration.
- dx indicates that x is the variable with respect to which
integration is to be performed.
The indefinite integral: f X . dX f X C
182
If f X = F(x) for all x in the domain of f
Integration Rules
Since the integration is the inverse of differentiation ,
then many rules for integration con be obtain by stating the
corresponding rules for differentiation.
Rule 1:
- If the function is: X n . dX
X n 1
Then : X . dX
n
C
n 1
So : ( C ) is a constaint and n 1
Where ( n ) is any rational number except ( n 1 ) and ( C )
is a constaint
Example : Find X 4 . dX
Solution
X 4 1
X 4 . dX C
4 1
X5
C
5
The derivative of the function is:
X5
Y C
5
1
Y X5 C
5
5 4
The first deriavative Y X X4
5
183
Rule 2:
If the function is: a. dx
Then : a. dX a X C
So: a is a real number constaint.
Example : Find the integral:
1- 10 . dX
1- 5. dQ
Solution
10 . dX 10X C
5. dQ 5 Q C
Rule 3:
If the function is: ax n . dx
X n 1
Then : aX n . dX a C
n 1
If (a) is real number and ( C ) is a constaint
Example : Find 4X 3 . dX
Solution
X4
4X 3 . dX 4 C
4
4X 3 . dX X 4 C
Example : Find 6X 2 . dX
184
Solution
X3
6X . dX 6 C
2
3
2X C
3
Rule 4:
If the function is: af X bg X
Or :
af X bg X eh X .... mL X
Then :
af X bg X a f X dX b g X dx
Or:
af X bg X eh X .... mL X
a f X b g X e h X .... m L X
This enables us to integrate an expression “ term by term
“ as the following example demonstrate.
Example : Find
2X 2 4X 6 dX
Solution
2X 2 4X 6 dX
2X 3 4X 7
3
7
2 3 4 7
X X
3 7
Example : Find
5X 2 3X 2 dX
185
Solution
5X 3X 2 dX
2
3
2
5X 3 3X 2
2X C
5 3 3 2
X X 2X C
3 2
Example : Find 10X 4 8X 3 3X 2 4X 10 dX
Solution
10 X 4 8 X 3 3 X 2 4 X 10 dX
5 4 3 2
10 X 8X 3X 4X
10 X C
5 4 3 2
2 X 5 2 X 4 X 3 2 X 2 10 X C
- The derivative of the function is:
Y 2 X 5 2 X 4 X 3 2 X 2 10 X C
The first dervaitive:
Y 10 X 4 8 X 3 3 X 2 4 X 10
Example : Find 20X 4 16X 3 12X 2 24X 7 dX
Solution
20 X 4 16 X 3 12 X 2 24 X 7 dX
20 X 5 16 X 4 12 X 3 24 X 2
7X C
5 4 3 2
4 X 5 4 X 4 4 X 3 12 X 2 7 X C
aX b . dX
n
Rule 5: If the function is:
So
186
aX b n 1
aX b . dX C
n
a n 1
1 aX b
n 1
C
a n 1
So: ( b , a ) are real numbers , n 1 and ( C ) is a
constant.
Example : Find 5 X 7 10 . dX
Solution
5 X 7 . dX
10
1 5 X 7
11
C
5 11
2 X 10 . dX
5
Example : Find
Solution
2 X 10 . dX
5
1 2 X 10
6
C
2 6
2 X 10
6
C
12
Rule 6: If the function is:
1
. dX
X
1
Then: . dX Ln X C
X
187
10
Example : Find . dX
10 X
Solution
10 1
. dX . dX Ln X C
10 X X
200
Example : Find . dX
200 X
Solution
200 1
. dX . dX Ln X C
200 X X
Result: If the numerator is the first derivative of the
denominator, the integration this function is the logarthame
function of denominator( L n ).
3X 2 4 X 15
Example : Find 3 . dX
X 2 X 2 15 X 8
Solution
the numerator is the first derivative of the denominator
3X 2 4 X 15
. dX
X 3 2 X 2 15 X 8
Ln X 3 2 X 2 15 X 8 C
8X 12
Example : Find . dX
X 2 3X 5
188
Solution
4 2 X 3
2 . dX
X 3X 5
2 X 3
4 2 . dX 4 Ln X 2 3 X 5 C
X 3X 5
Rule 7: If the function is:
e . dX e X C
X
1
e mX . dX e mX C
m
Example : Find e X 5 . dX
Solution
e X5 . dX e X5 C
Example : Find e 5 X 7 . dX
Solution
1 5 X 7
e5 X7 . dX e C
5
e3X 8 . dX
2
Example : Find
Solution
8 1 3X 2 8
e3X . dX C
2
e
6X
e3X 8
2
C
6X
189
Exmples on rules of integration.
- Find the indefinite integrals:
1. 4 X 3 . dX
2. 15 X 3 3 X 2 6 X 8 dX
3. 4 X 37 dX
4. 2 X 2 15 3 X 8dX
6
5. dX
X3
6. X dX
7. 36 X 28 X 5 dX
2
2X 4X 5X 10
20 X 15 X 10
3 2
8. dX
X X 2X
4 3
9. 7 e X dX
2
X
10. 5 4 X 3 dX
5
11. dX
X
Solution
4X 4
1. 4 X 3 . dX C X4 C
4
2. 15 X 3 3 X 2 6 X 8 dX
15 X 4 3X3 6X2
8X C
4 3 2
15 4
X X3 3X2 8X C
4
190
1 4 X 3
8
3. 4 X 3 dX C
7
4 8
4 X 3
8
C
32
4. 2 X 2 15 3 X 8dX
The multiple:
2 X 2 15 3 X 4 6 X3 8 X 2 45 X 60
6 X 3 8 X 2 45 X 60 dX
6 X 4 8 X 3 45 X 2
60 X C
4 3 2
6 8 45 2
X 4 X3 X 60 X C
4 3 2
1.5 X 4 2.76 X 3 22.5 X 2 60 X C
6
5. 3
dX 6 X 3 . dX
X
X 31
6 X 3 dX 6 C
3 1
X 2
6 3 X 2 C
2
3
2 C
X
6. X dX
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1
1 1
2
X
X dX
2
C
1
1
2
3
3
X2 2
C X2 C
3 3
2
3
2
2X
C
3
2
X3 C
3
6 X2 8 X 5
7. 3 dX
2 X 4 X 2
5 X 10
the numerator is the first derivative of the denominator
Ln 2X 3 4X 2 5X 10 C
20 X 3 15 X 2 10
8. dX
X 4
X 3
2 X
4 X 3 3 X 2 2
5 4 dX
X X 2X
3
4 X 3 3 X 2 2
5 4 dX
X X 2X
3
the numerator is the first derivative of the denominator
5 Ln X 4 X 3 2 X C
9. 7 e X dX
2
X
192
2
7 e X dX
X
2
7 e X . dX dX
X
1
7 e X dX 2 dX
X
e X
7X 2Ln X C
1
7e X 2 Ln X C
3
10. 5 4 X 3 dX 5 X 4 dX
3 7
1
4 4
5X 5X
C C
3 7
1
4 4
7
4 20 4 7
5 X 4 C X C
7 7
5
11. dX
X
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1
5
1
dX 5 X 2
dX
2
X
1
1 1
X 2
5 X 2
dX 5 C
1
1
2
1
5X 2
C
1
2
1 1
2 2
5 X C 10 X 2 C
1
10 X C
Calculate the constant of integration
We known the first derivative of a constant is equale zero.
For example :
- if function : Y X 4
the first derivative : Y 4 X 3
- if function : Y X 4 10
the first derivative : Y 4 X 3
- if function : Y X 4 20
the first derivative : Y 4 X 3
- if function : Y X 4 C where : C is constant
the first derivative : Y 4 X 3
We known :
194
4X 4
4 X . dX
3
C
4
X4 C
the integration of 4 X 3 . dX equal :
X4 1 or X 4 5 or X 4 10 or X4 C
when C is constant
if we known the first derivative f X
So: f X . dX f X C when C is
constant
We con calculate the value of c
Example : Y 6 X 2 20 X 600 dX
When y 20000 , and x 10 Find the value of C
Solution
Y 6 X 2 20 X 600 dX
6 X 3 20X 2
600 X C
3 2
y 2 3 10 2 600 C
Substutite the value of 20000 Y and 10 X
20000 2 10 10 10 600 10 C
3 2
20000 2000 1000 6000 C
2000 7000 C
C 20000 7000 13000
Y 2 X 3 10 X 2 600 X 13000
195
Definite integration
if the function a , b and the two points are Y f X , in
b
general , the definite integral f X . dX
a
.
denote the erea under the graph of f(x) between
x a and x b as shown in figure (1) the number a and b
are called the limit integration and it is assumed throughout
this section that a < b and that f(x) ≥ 0 as indicated in
figure (1) .
the technique of evaluating definite integrals is as
follows. A function F(x) is found which differentiates to f(x)
. the new function , F(x) , is then evaluated at the limits x = a
and x = b to get F(a) and F(b) . finally , the second numbers
is subtracted from the first to get the answer , F(b) – F(a)
Figure (1)
196
b
In terms of this new notation. to calculate f X . dX
a
,
first replace (x) by the upper limit (b) to obtain f(b), and
from this subtract f(a), obtained by letting x = a
b
f X . dX f b f a
a
- f b f X .dX xb
- f a f X . dX xa
- the constaint C drops out , this is always be the case.
Example : find 4X 3 . dX
4
solution
4
4X 4
4 4
4X . dX X4
3
3
3
4 3
4 3
4 4
256 81 175
Example : find 6 X 4 . dX
4
1
solution
197
4
6 X2
6 X 4 . dX
4
4X
1
2 1
3 X 2 4 X 1
4
3 4
4 4 3 1 4 1
2 2
48 16 3 4
64 1 65
Example : find X 2 X 1 . dX
4
solution
4
X3 X 2
4
X X 1 . dX
2
X
1 3 2 1
4
1 1
X3 X 2 X
3 2 1
1 3 1 2 1 3 1 2
4 4 4 1 1 1
3 2 3 2
1 1
64 8 4 0.50 1
1
3 2 3
21.33 0.83 20.5
198
2
Example : find X 1 . dX
0
solution
2
X2
2
1 2
2
0 X 1 . dX 2 X 2 X X
0 0
1 2 1 2
2 2 2 2 0
2
404
199
Exercises
1- Find the indefinite integrals:
3 X
2 X 5 . dX
2
1.
1
2. X 1
4X 5
3. 2 X 2 dX
5X 3
2- Find the indefinite integrals:
2
1) X
5 X dX
4X 8
2) 2 dX
X 4 X 3
3- Find the integration of these functions:
1) Y 6 x 2 20 x 600 dx
2) y 5 x 7 10 . dx
5
3) y dx
x
4) y 5 4 x3 dx
5 - Find the integration of these functions:
1. 2 X 35 dX
20 X 3 15 X 2 10
2. 4 dX
X X 10 X
3
3. 5 5 X 4 dX
3
4. dX
X
