8 techniques of integration

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8 TECHNIQUES OF INTEGRATION. - PowerPoint PPT Presentation

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Page 1: 8  TECHNIQUES OF INTEGRATION
Page 2: 8  TECHNIQUES OF INTEGRATION

Computer simulation of the Indonesian tsunami of December 26, 2004 (8 minutes after the earthquake), created using models of wave motion based on advanced calculus by Steven Ward, University of California at Santa Cruz.

8 TECHNIQUES OF INTEGRATIONIn Section 5.6 we introduced substitution, one of the most important techniques of integration. In this section, we develop a second fundamental technique, Integration by Parts, as well as several techniques for treating particular classes of functions such as trigonometric and rational functions. However, there is no surefire method, and in fact, many important antiderivatives cannot be expressed in elementary terms. Therefore, we discuss numerical integration in the last section. Every definite integral can be approximated numerically to any desired degree of accuracy.

Page 3: 8  TECHNIQUES OF INTEGRATION

The Integration by Parts formula is derived from the Product Rule:

' ' 'u x v x u x v x u x v x

is an antiderivative of the right-hand side.u x v x

' 'u x v x u x v x dx u x v x dx ' 'u x v x dx u x v x u x v x dx

Integration by Parts Formula

' 'u x v x dx u x v x u x v x dx

Subtract ' from both sides...u x v x dx

Page 4: 8  TECHNIQUES OF INTEGRATION

The Integration by Parts formula is derived from the Product Rule:

' ' 'u x v x u x v x u x v x

is an antiderivative of the right-hand side.u x v x

' 'u x v x u x v x dx u x v x dx ' 'u x v x dx u x v x u x v x dx

Integration by Parts Formula

' 'u x v x dx u x v x u x v x dx

Because the Integration by Parts formula applies to a product u(x)υ (x), we should consider using it when the integrand is a product of two functions.

'

Page 5: 8  TECHNIQUES OF INTEGRATION

cos ?x xdx Try writing as .cos 'x uvx

, ' cosu x x v x x

' 1, sinu x v x x

' 'u x v x dx u x v x u x v x dx cos sin sin sin

sin co

cos

s

x xdx x x xdx x x

x x C

C

x

x

Page 6: 8  TECHNIQUES OF INTEGRATION

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx?xxe dx

Try writing as .'xxe uv , ' xu x x v x e

' 1, xu x v x e

' 'u x v x dx u x v x u x v x dx

x x x x xxe dx xe e d e Cx x e

Here's some good advice:

Page 7: 8  TECHNIQUES OF INTEGRATION

2 2cos si 2 sinnx xdx x x x xdx , ' sinu x x v x x

' 1, cosu x v x x

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx

2 cos ?x xdx

Try writing the two factors of the integrand as '.uv

2 , ' cosu x x v x x ' 2 , sinu x x v x x

' 'u x v x dx u x v x u x v x dx

Integrating by Parts More Than Once

2 sin 2 c2 cos ocos s 2sinx x xdx xdx x x x Cx 2 2 sin 2 cos 2sincos x xx xdx x x x C

Page 8: 8  TECHNIQUES OF INTEGRATION

ln

' 1

' 1/

u x x

v x

u x x

v x x

Integration by Parts applies to definite integrals:

' 'b b

b

aa a

u x v x dx u x v x u x v x dx

3

1

ln ?xdx ????

3 33 3

1 11 1

ln ln 3l

3

n 3

3ln 3 3 1 ln 3 2

xdx x x dx x

' 1 is also used to find the antiderivative of inverse trig functions.v

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx

Try writing the two factors of the integrand as '.uv

' 'u x v x dx u x v x u x v x dx

Page 9: 8  TECHNIQUES OF INTEGRATION

sin

'

' cos

x

x

u x x

v x e

u x x

v x e

cos co sis nx x xe xdx e x e xdx sin con ssix x xe e x e xdxxdx

cos cos sin cosx x x xe xdx e x e x e xdx

cos

'

' sin

x

x

u x x

v x e

u x x

v x e

cos ?xe xdx

1 1co2 cos cos s sinsin cos2 2

x x x x x xe xdx e x e x e xd Cx e x e x

Integration by Parts applies to definite integrals:

' 'b b

b

aa a

u x v x dx u x v x u x v x dx

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx

Try writing the two factors of the integrand as '.uv

' 'u x v x dx u x v x u x v x dx

Page 10: 8  TECHNIQUES OF INTEGRATION

1

'

'

n

x

n

x

u x x

v x e

u x nx

v x e

3 3 23x x xx e dx x e x e dx

3 2

3 2

3 23 23 6 6

3 2

3 6

3 6 6x x x

x x x

x x

x

x

x

x

x e x e xe dx

x e x e xe e dx

x e x e xe e C x x ex C

1

3

Derive the reduction formula ,

then use it to evaluate .

n x n x n x

x

x e dx x e n x

e

e d

d

x

x x

1n x nx xn x e n ex xe x ddx

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx

Try writing the two factors of the integrand as '.uv

' 'u x v x dx u x v x u x v x dx

I think I seea pattern.

Page 11: 8  TECHNIQUES OF INTEGRATION