Integration of Nonlinear Equations of Mathematical Physics by Method of Inverse Scattering Method I
CHAPTER 4 INTEGRATION. Integration is the process inverse of differentiation process. The...
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Transcript of CHAPTER 4 INTEGRATION. Integration is the process inverse of differentiation process. The...
![Page 1: CHAPTER 4 INTEGRATION. Integration is the process inverse of differentiation process. The integration process is used to find the area of region under.](https://reader035.fdocuments.us/reader035/viewer/2022062301/56649e8f5503460f94b93d67/html5/thumbnails/1.jpg)
CHAPTER 4INTEGRATION
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Integration is the process inverse of diff erentiation process. The integration process is used to find the area of region under the curve.
INDEFINITE INTEGRAL
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Examples :
63 5 3 5 22
1. (2 ) 2 =-3 6
xx x dx x dx x dx x C
2. 3 3
= 3
x x
x
e dx e dx
e C
3
23. sin cos3
xx x dx x C
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EXERCISE 4
21
32
2/3
2
1. ( 3 6)
2. 3
3. 4 2cos3 6sin 2
4. 2
5. sin
x
x
x x dx
x dx
e x x dx
dx
x dx
2
2
6. 3
1 2 37.
2
68. 2 ln 2
3 29.
10. ( 2)( 1)
x
x
e dx
dxx x x
e dxx
x xdx
x
x x dx
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Examples :
THE DEFINITE INTEGRAL
22 32
1 1
1. 4 3 4 33
2 1 = 4 3(2) 4 3(1)
3 3
8 4 13 = 6 3
3 3 3
xx dx x C
C C
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Integration by Substitution – Change of Variables
Reversing the “chain rule” (from diff erentiation).General integration by substitution :
Integration by Substitution steps:
Figure out the “inner” function; cal l i t u(x) . Compute . . Replace al l expressions involving the variable x and dx with the new variable u and du . Use the diff erential formula to replace the diff erential dx .
Evaluate the result ing “u” integral. I f you can’t evaluate the integral, try a diff erent choice of u .
Replace al l occurrences of the variable u in the antiderivative with the appropriate function of x .
TECHNIQUES OF INTEGRATION
'( ( )). '( ) ( ( ))f g x g x dx f g x C
du
dx
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Example 1 : Calculate 3 4x dx
1 2 3 2 3 2
3 4,
3
1
31 1 1 2 2
(3 4)3 3 3 3 9
Let u x
du
dx
dx du
u du u du u C x C
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Example 2:Find
43 2xx e dx
4
4
4
3
3
3 2 33
2
Let 2
4
4
41
41 1
= 4 4
x u
u
u x
u x
dux
dxdu
dxx
dux e dx x e
x
e du
e C e C
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EXERCISE 5
2 3
5 2 5 2
65
1. ans: 1 ln 11
ln ln2. ans:
31
3. ans: 5
14. ans: ln 1
1
2 65. 2 6 ans:
12
6. ans: ln
x x
xx
x xx x
x x
xdx x x C
x
x xdx C
x
e dx e C
dx e Ce
xx dx C
e edx e e C
e e
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43
3
2
5 3
2 2
2
sin( )7. sin( ) cos ans:
41
8. 2 7 ans: 2 731 1
9. 1 2 ans: 1 2 1 210 6
sin(tan )10. ans: cos(tan )
cos
xx xdx C
x dx x C
x xdx x x C
d C
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Integration by Parts
Example 3 :Find
udv uv vdu 2 ln x x dx
2
32
3 3
3 2
3 3
3 3
ln
1
3
ln3 3
1 1 = ln
3 3
1 = ln
3 3 3
= ln3 9
u x dv x
du xv x dx
dx xdx
dux
x x dxuv vdu x
x
x x x dx
x xx C
x xx C
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EXERCISE 6
2 22
3 5
2 2
22
1. ans: 2 4
2 42. 5 ans: 5 5
3 15
3. sin ans: cos sin
cos10 cos104. sin(10 ) ans: - sin10
10 50 500
5. sin
x xx xe e
xe dx C
xx x dx x x C
x xdx x x x C
x x x xx x x C
x
32 3
2 2
1 ans: cos sin
2
ln 16. ln ans:
3 9
7. ans: 2 2
x x
x x
e dx e x x C
x xx xdx x C
x e dx e x x C
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Partial FractionsA quotient of polynomials : can be expressed as a simpler fraction called partial fractions.
This technique is used for rewriting problems so that they can integrate.
For example, the integral can be rewritten as using the method partial fractions. Then it is easily integrated as
( )( )
( )
P xf x
Q x
2
7
6
xdx
x x
2 1( )
3 2dx
x x
2ln 3 ln 2x x C
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Types of Partial Fraction
1. Improper Fraction
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2. Proper Fraction
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Example 4: (Case 1)5
( 2)( 3) 2 3
( 3) ( 2) =
( 2)( 3) ( 2)( 3)
( 3) ( 2) 5
3 2 5
( ) 3 2 5
0
3 2 5
1
1
5 1 1ln 2 ln 3
( 2)( 3) 2 3
A Bdx
x x x x
A x B x
x x x x
A x B x
Ax A Bx B
x A B A B
A B
A B
A
B
dx dx x x Cx x x x
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Example 5 :(Improper Fraction)3 2
2 2
22
2
2
2
5
6 61 5
= 2 61 5
= 2 ( 2)( 3)
1 =
2 ( 2) ( 3)
1 =
2
x x x xdx x dx
x x x xx
x dxx x
xx dx
x x
A Bx dx
x x
x
2
2
( 3) ( 2)
( 2) ( 3)
1 2 3 =
2 ( 2) ( 3)
1 = 2ln 2 3ln 3
2
A x B xdx
x x
x dxx x
x x x C
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EXERCISE 7
6
2
2 2
2
3
7 11. ans: 5ln 3 2ln 1
( 3)( 1)
1 3 12. ans: 2 6
( 2) 12
3 53. ans: 2 3ln 1
1 2
3 34.
( 1)
xdx x x C
x x
xdx x C
x x
x x xdx x x C
x
xdx
x
22
3 2
2
2
3 2
6 ans: 3ln 1
1
5 10 15. ans: 2ln 2 5
2 5 21
6. ans: ln 2 ln 24 4
17. ans: ln
x Cx
x xdx x x x C
x x xdx
x x Cx
x xdx x
x x x
1
2C