ESSENTIAL CALCULUS CH06 Techniques of integration

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ESSENTIAL CALCULUSESSENTIAL CALCULUS

CH06 Techniques of CH06 Techniques of integrationintegration

In this Chapter:In this Chapter:

6.1 Integration by Parts

6.2 Trigonometric Integrals and Substitutions

6.3 Partial Fractions

6.4 Integration with Tables and Computer Algebra Systems

6.5 Approximate Integration

6.6 Improper Integrals

Review

Chapter 6, 6.1, P307

x dxxfxgxgxfdxxgxf )(')()()()(')(

Chapter 6, 6.1, P307

vduuvudv

Chapter 6, 6.1, P309

dxxfxgxgxfdxxgxf ba

ba

ba )(')()]()()(')(

Chapter 6, 6.2, P314

▓How to Integrate Powers of sin x and cos xFrom Examples 1– 4 we see that the following strategy works:

Chapter 6, 6.2, P314

(i) If the power of cos x is odd, save one cosine factor and use cos2x=1-sin2x to express the remaining factors in terms of sin x. Then substitute u=sin x.

Chapter 6, 6.2, P314

(ii) If the power of sin x is odd, save one sine factor and use sin2x=1-cos2x to express the remaining factors in termsof cos x. Then substitute u=cos x.

Chapter 6, 6.2, P314

(iii) If the powers of both sine and cosine are even, use the half-angle identities:

It is sometimes helpful to use the identity

)2cos1(2

1sin 2 x

)2cos1(2

1cos2 x

xxx 2sin2

1cossin

Chapter 6, 6.2, P315

▓How to Integrate Powers of tan x and sec xFrom Examples 5 and 6 we have a strategy for two cases

Chapter 6, 6.2, P315

(i) If the power of sec x is even, save a factor of sec2x and use sec2x=1+tan2x to express the remaining factors in terms of tan x.Then substitute u=tan x.

Chapter 6, 6.2, P315

(ii) If the power of tan x is odd, save a factor of sec x tan x and use tan2x=sec2x-1 to express the remaining factors in terms of sec x. Then substitute u sec x.

Chapter 6, 6.2, P315

Cxxdx seclntan

Chapter 6, 6.2, P315

Cxxxdx tanseclnsec

Chapter 6, 6.2, P317

TABLE OF TRIGONOMETRIC SUBSTITUTIONS

Expression Substitution Identity

22 xa

22 xa

22 ax

22,sin

ax

22,tan

ax

2

3

20,sec

orax

22 cossin1

22 sectan1

22 tan1sec

Ca

x

aax

dx

1

22tan1

Chapter 6, 6.3, P326

Chapter 6, 6.5, P336

Chapter 6, 6.5, P336

Chapter 6, 6.5, P336

Chapter 6, 6.5, P336

If we divide [a,b] into n subintervals of equal length ∆x=(b-a)/n , then we have

where X*1 is any point in the ith subinterval

[xi-1,xi].

n

ii

ba xxfdxxf

1

*)()(

Chapter 6, 6.5, P336

Left endpoint approximation

n

ii

ba xxfLndxxf

11)()(

Chapter 6, 6.5, P336

Right endpoint approximation

n

ii

ba xxfRndxxf

1

)()(

Chapter 6, 6.5, P336

MIDPOINT RULE

)]()(([)( 21 nba xf‧‧‧xfxfxMndxxf

wheren

abx

and ],[int)(2

1111 iiii xxofmidpoxxx

Chapter 6, 6.5, P337

Chapter 6, 6.5, P337

TRAPEZOIDAL RULE

)]()(2)(2)(2)([2

)( 1210 nnba xfxf‧‧‧xfxfxf

xTndxxf

Where ∆x=(b-a)/n and xi=a+i∆x.

Chapter 6, 6.5, P337

errorionapproximatdxxfba )(

Chapter 6, 6.5, P339

3. ERROR BOUNDS Suppose │f”(x)│≤K for a≤x≤b. If ET and EM are the errors in the Trapezoidal and Midpoint Rules, then

2

3

12

)(

n

abKET

2

3

24

)(

n

abKEM

and

Chapter 6, 6.5, P340

Chapter 6, 6.5, P340

Chapter 6, 6.5, P342

SIMPSON’S RULE

‧‧‧xfxfxfxfx

Sndxxfba

)(4)(2)(4)([3

)( 3210

)]()(4)(2 12 nnn xfxfxf

Where n is even and ∆x=(b-a)/n.

Chapter 6, 6.5, P343

ERROR BOUND FOR SIMPSON’S RULE Suppose that │f(4)(x)│≤K for a≤x≤b. If Es is the error involved in using Simpson’s Rule, then

4

5

180

)(

n

abKEs

Chapter 6, 6.6, P347

In defining a definite integral we dealt with a function f defined on a finiteinterval [a,b]. In this section we extend the concept of a definite integral to the casewhere the interval is infinite and also to the case where f has an infinite discontinuityin [a,b]. In either case the integral is called an improper integral.

dxxfba )(

Chapter 6, 6.6, P347

Improper integrals:

Type1: infinite intervals

Type2: discontinuous integrands

DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 1(a) If exists for every number t≥a, then

provided this limit exists (as a finite number).(b) If exists for every number t≤b, then

provided this limit exists (as a finite number).

dxxfdxxf taa )()( lim

1

dxxfbt )(

dxxfta )(

dxxfdxxf bt

b )()( lim1

Chapter 6, 6.6, P348

Chapter 6, 6.6, P348

The improper integrals and are called convergent if the corresponding limit exists and divergent if the limit does not exist.(c) If both and are convergent, then we define

In part (c) any real number can be used (see Exercise 52).

dxxfa )( dxxfb )(

dxxfa )( dxxfa )(

dxxfdxxfdxxf aa )()()(

Chapter 6, 6.6, P351

dxx p1

1 is convergent if p>1 and divergent if p≤1.

Chapter 6, 6.6, P351

3.DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 2(a)If f is continuous on [a,b) and is discontinuous at b, th

en

if this limit exists (as a finite number).(b) If f is continuous on (a,b] and is discontinuous at a, t

hen

if this limit exists (as a finite number).

dxxfdxxf ta

b

ba )(lim)(

1

dxxfdxxf bt

a

ba )(lim)(

1

Chapter 6, 6.6, P351

The improper integral is called convergent if the corresponding limit exists and divergent if the limit does not exist

(c)If f has a discontinuity at c, where a<c<b, and both and are convergent, then we define

dxxfba )(

dxxfca )( dxxfbc )(

dxxfdxxfdxxf bc

ca

ba )()()(

Chapter 6, 6.6, P352

erroneous calculation:

This is wrong because the integral is improper and must be calculated in terms of limits.

2ln1ln2ln1ln1

30

30

xx

dx

Chapter 6, 6.6, P353

COMPARISON THEOREM Suppose that f and g are continuous functions with f(x)≥g(x)≥0 for x≥a .

(b) If is divergent, then is divergent.(a) If is convergent, then is convergent.

gxxfa )(

fxxfa )( gxxfa )(

fxxfa )(