chapter 7: integration techniques, l’hôpital’s rule, and improper integrals
TRANSCRIPT
Chapter 7: Integration Techniques, L’Hôpital’s Rule, and
Improper Integrals
Section 7.2: Integration by Parts
• Particularly useful for integrals involving products of algebraic and transcendental functions– Examples:
• Based on the formula for the derivative of a product:
uvvu
dx
duv
dx
dvuuv
dx
d
Section 7.2: Integration by Parts• If u' and v' are both continuous, this can be
integrated to obtain:
• This yields Theorem 7.1:
• Guidelines for Integration by Parts:1. In order to find u follow the acronym LIPET.2. Once u is identified the rest of the integral will be dv.
vduudv
dxuvdxvuuv
vduuvudv
Section 7.2: Integration by Parts
Example 1 Solve xdx3ln using integration by parts
SOLUTION
Step 1: Break equation into vduuvudv ;
xv
dxx
du
dxdv
xu
3
3ln
Step 2: Set up equation:
dxx
xxxxdx
33ln3ln
Simplify. Cxxx 33ln
Section 7.2: Integration by PartsExample 3: Solve dxex x
1
0
2 using integration by parts.
SOLUTION
Step 1: Break equation into vduuvudv ; x
x
evxdxdu
dxedvxu
2
2
Step 2: Set up equation:
dxxeexdxex xxx 1
0
21
0
2 2
Uh oh! We need to do another integration by parts!
Step 3: Break equation into vduuvudv ; x
x
evdxdu
dxedvxu
2
2
Step 4: Set up equation and solve.
)22(1
0
21
0
2 dxexeexdxex xxxx
2
222
221
02
e
eee
exeex xxx
Section 7.3: Trigonometric Integrals
• This section helps to evaluate integrals of the forms:
• The following identities can be helpful:
xdxx nm cossin xdxx nm tansec
2
2cos1cos
2
2cos1sin
1cossin
2
2
22
xx
xx
xx
Section 7.3: Trigonometric Integrals• Guidelines for evaluating integrals involving
sine and cosine (refer to p. 490):1. If the power of the sine is odd and positive, save
one sine factor and convert the rest to cosine.
2. If the power of the cosine is odd and positive, save one cosine factor and convert the remaining factors to sine.
3. If the powers of both sine and cosine are even and nonnegative, make repeated use of the previously-mentioned identities.
Section 7.3: Trigonometric Integrals
Example 1 Solve xdxx sincos3
SOLUTION
Using integral properties we know xdxsin is Ccos
Multiply the integral by -1:- xdxx sincos3
Integrate: Cx
4
cos4
Section 7.3: Trigonometric Integrals
Example 2 Solve xdxx 2cos2sin 5
SOLUTION
Using integral properties we know xdxcos is Cx sin
xdxx 2cos2sin 5
Take out ½ for the 2x: xdxx 2cos2sin2
1 5
Integrate: Cx
6
2sin
2
1 6
Multiply: 12
2sin 6 x+C
Section 7.4: Trigonometric Substitution
• This helps to solve integrals involving the radicals:
• Uses the Pythagorean identities:
222222 ;; auuaua
22
22
22
csccot1
sectan1
1cossin
• For integrals involving
let u = a sin θ– Then
• For integrals involving
let u = a tan θ– Then
• For integrals involving
let u = a sec θ– Then
Section 7.4: Trigonometric Substitution
22 au
22 ua
22 ua
θaua cos22
θaua tan22
sec22 aau
Section 7.4: Trigonometric SubstitutionExample 1
dxx
x
92
Determine u and a
Since 92 x follows the format of 22 ua tanau therefore tanu
22 ua = seca tanx ddx 2sec Substitute:
sec9
sectan 2 d
dsectan9
Integrate: sec9
Substitute Back:
Cx 92
Section 7.4: Trigonometric SubstitutionExample 2
dxx
9
12
Determine u and a
Since 92 x follows the format of 22 au secau therefore sec3u
9tan 2 x
sec33
sec
x
x
ddx tansec3 Substitute:
d
d
sec3
sec
tansec3
Integrate: C tan3sec3ln
Substitute back:
Cxx 9ln 2
Section 7.5: Partial Fractions
• Allows a rational function to be decomposed into simpler rational functions, to which basic integration formulas can be applied
• Methods of Decomposing N(x)/D(x):1. Divide if numerator is greater than denominator
2. Factor the denominator
3. Use linear factors
4. Use quadratic factors
Section 7.5: Partial FractionsExample 1
Evaluate
)4)(53(
18
xx
x
SOLUTION
Find A and B: )4)(53(
18
453
xx
x
x
B
x
A
Multiply by the product of the denominators: 18)153()4( xxBxA
Simplify: 18)54()3(
18534
xBAxBA
xBBxAAx
Solve for A and B:
2;7
1834
13
BA
BA
BA
Rewrite the integral using this information: dxx
dxx
4
2
53
7
Solve: Cxx 4ln23
5ln
3
7x
Section 7.5: Partial FractionsExample 2
Evaluate 2)1(
42
x
x
SOLUTION
Find A and B: 22 )1(
42
)1(1
x
x
x
B
x
A
Multiply all terms by 2)1( x
Simplify:
42)(
42
42)1(
xABAx
xBAAx
xBxA
Therefore: 6;4;2 BABA
Rewrite integral: dxx
dxx
2)1(
6
1
2
Solve: Cx
x
1
61ln2
Section 7.7: Indeterminate Forms and L’Hôpital’s Rule
• The forms 0/0 and ∞/ ∞ are called indeterminate because they do not guarantee that a limit exists, if one does exist.
• L’Hôpital’s Rule helps to solve these types of problems:
• Therefore, finding the limit of the derivatives allows one to find the limit of an indeterminate form.
)('
)('lim
)(
)(lim
xg
xf
xg
xfcxcx
Section 7.7: Indeterminate FormsExample 1
Find x
xxx
sin2lim
0
SOLUTION
Because plugging in 0 yields the indeterminate form 0
0, we use L’Hôpital’s Rule and
take the derivative of both the numerator and the denominator.
1
cos2lim
0
xx
The limit, therefore, equals 1.
Section 7.7: Indeterminate FormsExample 2
Find 20 2
424
limx
xx
x
.
SOLUTION Because this is indeterminate, we take the derivative of both the top and bottom:
xx
x 44
1
42
1
lim0
This form is still indeterminate, so we apply L’Hôpital’s Rule again, yielding:
128
1
4
)4(4
1
lim
2
3
0
x
x
Section 7.8: Improper Integrals
• The definition of a proper integral
requires that the interval [a, b] be finite.
• If either or both of the limits of an integral are infinite, or if f has a finite number of infinite discontinuities in the interval [a, b], then the integral is improper.
b
a
dxxf )(
Section 7.8: Improper Integrals
• A function f is said to have an infinite discontinuity at c if, from the right or left:
or
)(lim cfcx
)(lim cfcx
Section 7.8: Improper Integrals
• Definition of Improper Integrals with Infinite Integration Limits:
1. If f is continuous on the interval then
2. If f is continuous on the interval then
3. If f is continuous on the interval then
,a
a
b
ab
dxxfdxxf )(lim)(
b,
b b
aa
dxxfdxxf )(lim)(
,
c
c
dxxfdxxfdxxf )()()(
Section 7.8: Improper Integrals
Example 1
Determine convergence or divergence of the integral
12x
dx
SOLUTION
Step 1: Put into proper form
a
a x
dx
x
dx
12
12
lim
Step 2: Solve and determine behavior.
11
11limlim
12
ax
dxa
a
a (converges)
Section 7.8: Improper IntegralsExample 2
Solve the improper integral
2
03 1x
dx
SOLUTION Step 1: Split into two integrals that show the number that makes the equation undefined:
2
13
1
03
2
03 111 x
dx
x
dx
x
dx
Step 2: Convert to proper integrals:
2
31
031
2
13
1
03 1
lim1
lim11 a
a
a
ax
dx
x
dx
x
dx
x
dx
Step 3: Solve.
02
3
2
3
2
31
2
3lim
1lim
2
31
2
3lim
1lim
2
3
2
1
2
31
0
3
2
1
031
aa
a
a
a
a
a
a
xx
dx
xx
dx