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ESSENTIAL CALCULUS CH06 Techniques of integration. 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 - PowerPoint PPT Presentation
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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 )(