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Math 106: Integration Review - Solutions, Exams of Calculus

Solutions to the integration review questions for math 106, including tips on integration techniques, examples of integration using substitution, partial fractions, and trigonometric functions, and improper integrals.

Typology: Exams

2012/2013

Uploaded on 03/16/2013

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Math 106: Review for Exam II - SOLUTIONS
INTEGRATION TIPS
Substitution: usually let u= a function that’s “inside” another function, especially if du (possibly off
by a multiplying constant) is also present in the integrand.
Parts: Zudv=uv Zvdu or Zuv0dx =uv Zu0vdx
How to choose which part is u? Let ube the part that is higher up in the LIATE mnemonic below.
(The mnemonics ILATE and LIPET will work equally well if you have learned one of those instead;
in the latter Ais replaced by P, which stands for “polynomial”.)
Logarithms (such as ln x)
Inverse trig (such as arctan x, arcsinx)
Algebraic (such as x, x2,x
3+4)
Trig (such as sin x, cos 2x)
Exponentials (such as ex,e
3x)
Rational Functions (one polynomial divided by another): if the degree of the numerator is greater than
or equal to the degree of the denominator, do long division then integrate the result.
Partial Fractions: here’s an illustrative example of the setup.
3x2+11
(x+ 1)(x3)2(x2+5) =A
x+1+B
x3+C
(x3)2+Dx +E
x2+5
Each linear term in the denominator on the left gets a constant above it on the right; the squared
linear factor (x3) on the left appears twice on the right, once to the second power. Each irreducible
quadratic term on the left gets a linear term (Dx +Ehere) above it on the right.
Trigonometric Substitutions: some suggested substitutions and useful formulae follow.
Radical Form pa2x2pa2+x2px2a2
Substitution x=asin t x =atan t x =asec t
sin2x+ cos2x= 1 tan2x+ 1 = sec2x
sin2x=1
2cos(2x)
2cos2x=1
2+cos(2x)
2
sin(2x) = 2 sin xcos x
Powers of Trigonometric Functions: here are some strategies for dealing with these.
Zsinmxcosnxdx Possible Strategy Identity to Use
modd Break off one factor of sin xand substitute u= cos x. sin2x=1cos2x
nodd Break off one factor of cos xand substitute u= sin x. cos2x=1sin2x
m, n even Use sin2x+ cos2x= 1 to reduce to only powers of sin xsin2x=1
2cos(2x)
2
or only powers of cosx, then use integration by parts cos2x=1
2+cos(2x)
2
or identities shown to right of this box.
pf3
pf4
pf5

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Math 106: Review for Exam II - SOLUTIONS

INTEGRATION TIPS

  • Substitution: usually let u = a function that’s “inside” another function, especially if du (possibly off

by a multiplying constant) is also present in the integrand.

  • Parts:

u dv = uv −

v du or

uv

′ dx = uv −

u

′ v dx

How to choose which part is u? Let u be the part that is higher up in the LIATE mnemonic below.

(The mnemonics ILATE and LIPET will work equally well if you have learned one of those instead;

in the latter A is replaced by P, which stands for “polynomial”.)

Logarithms (such as ln x)

Inverse trig (such as arctan x, arcsin x)

Algebraic (such as x, x

2 , x

3

Trig (such as sin x, cos 2x)

Exponentials (such as e

x , e

3 x )

  • Rational Functions (one polynomial divided by another): if the degree of the numerator is greater than

or equal to the degree of the denominator, do long division then integrate the result.

Partial Fractions: here’s an illustrative example of the setup.

3 x

2

  • 11

(x + 1)(x − 3) 2 (x 2

A

x + 1

B

x − 3

C

(x − 3) 2

Dx + E

x 2

  • 5

Each linear term in the denominator on the left gets a constant above it on the right; the squared

linear factor (x − 3) on the left appears twice on the right, once to the second power. Each irreducible

quadratic term on the left gets a linear term (Dx + E here) above it on the right.

  • Trigonometric Substitutions: some suggested substitutions and useful formulae follow.

Radical Form

a 2 − x 2

a 2

  • x 2

x 2 − a 2

Substitution x = a sin t x = a tan t x = a sec t

sin

2 x + cos

2 x = 1 tan

2 x + 1 = sec

2 x

sin

2 x =

cos(2x)

cos

2 x =

cos(2x)

sin(2x) = 2 sin x cos x

  • Powers of Trigonometric Functions: here are some strategies for dealing with these.

sin

m x cos

n x dx Possible Strategy Identity to Use

m odd Break off one factor of sin x and substitute u = cos x. sin

2 x = 1 − cos

2 x

n odd Break off one factor of cos x and substitute u = sin x. cos

2 x = 1 − sin

2 x

m, n even Use sin

2 x + cos

2 x = 1 to reduce to only powers of sin x sin

2 x =

cos(2x)

or only powers of cos x, then use integration by parts cos

2 x =

cos(2x)

or identities shown to right of this box.

tan

m x sec

n x dx Possible Strategy Identity to Use

m odd Break off one factor of sec x tan x and substitute u = sec x. tan

2 x = sec

2 x − 1

n even Break off one factor of sec

2 x and substitute u = tan x. sec

2 x = tan

2 x + 1

m even, n odd Use identity at right to reduce to powers of sec x alone. tan

2 x = sec

2 x − 1

Then use integration by parts or reduction formula (if allowed).

Useful Trigonometric Derivatives and Antiderivatives

d

dx

tan x = sec

2 x

d

dx

sec x = sec x tan x

sec x dx = ln | sec x + tan x| + C

  • Improper integrals: look for ∞ as one of the limits of integration; look for functions that have a vertical

asymptote in the interval of integration. It may be useful to know the following limits.

lim x→∞

e

x = ∞

lim x→∞

e

−x = 0 Note: this is the same as lim x→−∞

e

x .

lim x→∞

1 /x = 0 Note: the answer is the same for lim x→∞

1 /x

2 and similar functions.

lim x→ 0

1 /x = ∞ Note: the answer is the same for lim x→ 0

1 /x

2 and similar functions.

lim x→∞

ln x = ∞

lim x→ 0 +

ln x = −∞

lim x→∞

arctan x = π/ 2

  1. Evaluate the following.

(a) Let u = sin x, so du = cos x dx.

sin

6 x cos

3 x dx =

sin

6 x(1 − sin

2 x) cos x dx Use cos

2 x = 1 − sin

2 x.

u

6 (1 − u

2 ) du

(u

6 − u

8 ) du

u 7

u 9

+ C

sin

7 x

sin

9 x

+ C

(b) Let x = 10 tan t, so dx = 10 sec

2 t dt.

x

y

t

x

2

  • 10

2 = y

2 ⇒ y =

x 2

  • 100

sec t =

hyp

adj

x^2 + 100

tan t =

opp

adj

x

3 x

2

  • 2x − 13 = (Ax + B)(x − 3) + C(x

2

  • 1).

Let x = 3. Then 20 = C(10), so C = 2.

Let x = 0. Then −13 = B(−3) + 2(1), so B = 5.

Let x = 1. Then −8 = (A(1) + 5)(−2) + 2(2), so A = 1.

3 x

2

  • 2x − 13

(x − 3)(x 2

dx =

∫ [^

x + 5

x 2

  • 1

x − 3

]

dx

∫ [^

x

x 2

  • 1

x 2

  • 1

x − 3

]

dx Let u = x

2

  • 1, so du = 2xdx.

2 du

u

∫ [^

x 2

  • 1

x − 3

]

dx

ln u

  • 5 arctan x + 2 ln |x − 3 | + D

ln(x 2

  • 5 arctan x + 2 ln |x − 3 | + D

(f) Since the degree of the numerator is greater than or equal to the degree of the denominator, we

do long division.

4 x

2 − 3 x + 2 +

x − 6

x − 6

4 x

3 − 27 x

2

  • 20x − 17

4 x

3 − 24 x

2

− 3 x

2

− 3 x

2 +18x

2 x

2 x − 12

Now, we compute the integral. ∫ 4 x

3 − 27 x

2

  • 20x − 17

x − 6

dx =

∫ [

4 x

2 − 3 x + 2 −

x − 6

]

dx =

4 x

3

3 x

2

  • 2x − 5 ln |x − 6 | + C

(g) This integral is improper at x = 1 because the integrand has a vertical asymptote there.

1

x − 1

dx = lim t→ 1 +

t

x − 1

dx

= lim t→ 1 +

ln |x − 1 |

3

t

= lim t→ 1 +

[ln | 3 − 1 | − ln |t − 1 |]

Since lim t→ 1 +

(− ln |t − 1 |) = ∞, this integral diverges (to ∞).

  1. Find the second-degree Taylor polynomial for f(x) =

f(x) = x

f(x) = x

x centered at xxx = 100= 100= 100.

f(x) = x

1 / 2 f(100) = 10

f

′ (x) =

x

− 1 / 2

x

f

′ (100) =

f

′′ (x) =

x

− 3 / 2

4 x 3 / 2

f

′′ (100) =

3 / 2

P 2 (x) = f(100) + f

′ (100)(x − 100) +

f ′′ (100)

(x − 100)

2

x − 100

(x − 100)

2

  1. What is the maximum possible error that can occur in your Taylor approximation from

the previous problem on the interval [100, 110]?

We know that |f(x) − Pn(x)| ≤

Kn+

(n + 1)!

|x − x 0 |

n+ .

In this case, n = 2, x 0 = 100, and x = 110 (the farthest from x 0 that we are considering).

K 3 = max of |f

′′′ (x)| on [100, 110] = max of |

8 x 5 / 2

| on [100, 110] =

5 / 2

Putting this all together, we have |f(x) − P 2 (x)| ≤

3 800 , 000

3!

3

  1. Use comparisons to show whether each of the following converges or diverges. If an

integral converges, also give a good upper bound for its value.

(a)

1

6 + cos x

x

  1. 99

dx

For all x ≥ 1, we have

6 + cos x

x

  1. 99

x

  1. 99

x

  1. 99

because the minimum value of cos x is −1.

Since

1

5 dx

x

  1. 99

diverges (compute yourself or notice that p = 0. 99 < 1), we know that the integral

in question must diverge too.

(b)

1

4 x

3 − 2 x

2

2 x^4 + x^5 + 1

dx

For all x ≥ 1, we have

4 x

3 − 2 x

2

2 x 4

  • x 5
  • 1

4 x

3

x 5

x 2

. (We’ve made the denominator smaller and

the numerator larger, so the new fraction is larger.)

1

dx

x 2

= 4 lim t→∞

t

1

dx

x 2

= 4 lim t→∞

x

t

1

= 4 lim t→∞

[

t

]

= 4[0 − (−1)]

Therefore, the original integral in question must converge to a value less than 4.

  1. The probability density function (pdf ) of the length (in minutes) of phone calls on a

certain wireless network is given by f(x) = ke

− 0. 2 x f(x) = ke

− 0. 2 x f(x) = ke

− 0. 2 x where xxx is the number of minutes.

Note that the domain is xxx ≥≥≥ 000 since we can’t have a negative number of minutes.

(a) What must be the value of kkk?