Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Final Exam Review Key - Calculus II | MATH 132, Exams of Calculus

Material Type: Exam; Class: Calculus II; Subject: Mathematics; University: Saint Mary's College; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 08/05/2009

koofers-user-p0g
koofers-user-p0g 🇺🇸

10 documents

1 / 8

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Final Exam Review Key–Math 132
1. a) Integration by Parts: Let u= ln(x4) = 4 ln x,du = (4/x)dx,dv =x2dx, and v=(1/x)dx.
Zx2·ln(x4)dx =4 ln x
x+Zx2dx =4 ln x
x4
x+C
b) Integration by Parts, twice: Let u=x2,du = 2xdx,dv =exdx,v=ex.
Zx2exdx =x2ex2Zxexdx
Now, let u=x,du =dx,dv =exdx, and v=ex. Then
Zx2exdx =x2ex2xexZexdx=x2ex2xex+ 2ex+C.
c) Double-angle Formula:
Zsin2(3x)dx =Z1
21
2cos(6x)dx =x
2sin(6x)
12 +C
d) U-substitution: Let u= tan x,du = sec2xdx. Then
Ztan4x(tan2x+ 1) sec2xdx =Zu4(u2+ 1)du =u7
7+u5
5+C=tan7x
7+tan5x
5+C.
e) U-substitution: Let u=ex,du =exdx. Then
Zexsin(ex)dx =Zsin udu =cos u+C=cos(ex) + C.
f) Integration by Parts: Let u= arctan x,du =dx
1+x2,dv =dx,v=x. Then
Zarctan xdx =xarctan xZx
1 + x2dx =xarctan x1
2ln(1 + x2) + C.
g) U-substitution: Let u=ex,du =exdx. Then
Zexdx
1e2x=Zdu
1u2= arcsin u+C= arcsin(ex) + C.
h) Partial Fraction Decomposition:
Z3x5
(x+ 1)2(x2)dx =ZA
x+ 1 +B
(x+ 1)2+C
x2dx
1
pf3
pf4
pf5
pf8

Partial preview of the text

Download Final Exam Review Key - Calculus II | MATH 132 and more Exams Calculus in PDF only on Docsity!

Final Exam Review Key–Math 132

  1. a) Integration by Parts: Let u = ln(x^4 ) = 4 ln x, du = (4/x)dx, dv = x−^2 dx, and v = −(1/x)dx. ∫ x−^2 · ln(x^4 )dx = −4 ln x x

x−^2 dx = −4 ln x x

x

+ C

b) Integration by Parts, twice: Let u = x^2 , du = 2xdx, dv = exdx, v = ex. ∫ x^2 exdx = x^2 ex^ − 2

xexdx

Now, let u = x, du = dx, dv = exdx, and v = ex. Then

∫ x^2 exdx = x^2 ex^ − 2

xex^ −

exdx

= x^2 ex^ − 2 xex^ + 2ex^ + C.

c) Double-angle Formula: ∫ sin^2 (3x)dx =

cos(6x)dx = x 2

sin(6x) 12

+ C

d) U-substitution: Let u = tan x, du = sec^2 xdx. Then ∫ tan^4 x(tan^2 x + 1) sec^2 xdx =

u^4 (u^2 + 1)du = u^7 7

u^5 5

+ C =

tan^7 x 7

tan^5 x 5

+ C.

e) U-substitution: Let u = ex, du = exdx. Then ∫ ex^ sin(ex)dx =

sin udu = − cos u + C = − cos(ex) + C.

f) Integration by Parts: Let u = arctan x, du = (^) 1+dxx 2 , dv = dx, v = x. Then ∫ arctan xdx = x arctan x −

x 1 + x^2 dx = x arctan x −

ln(1 + x^2 ) + C.

g) U-substitution: Let u = ex, du = exdx. Then ∫ exdx √ 1 − e^2 x^

du √ 1 − u^2

= arcsin u + C = arcsin(ex) + C.

h) Partial Fraction Decomposition: ∫ 3 x − 5 (x + 1)^2 (x − 2) dx =

A

x + 1

B

(x + 1)^2

C

x − 2 dx

9 x + 1

− 2 3 (x + 1)^2

11 9 x − 2 dx =

ln |x + 1| +

3(x + 1)

ln |x − 2 | + C

i) Partial Fraction Decomposition: ∫ dx (2x − 3)(2x + 3)

dx 2 x − 3

dx 2 x + 3

ln | 2 x − 3 | −

ln | 2 x + 3| + C

j) Simplify: ∫ x + 5 x^2 + 9 dx =

2 x x^2 + 9 dx + 5

dx x^2 + 9

ln(x^2 + 9) +

arctan( x 3

) + C

k) U-substitution: Let u = 9 − x^2 , du = − 2 xdx. Then ∫ √xdx 9 − x^2

u−^1 /^2 du = −u^1 /^2 + C = −

9 − x^2 + C.

l) Trig Substitution: Let x = 3 sin θ, dx = 3 cos θdθ. Then ∫ dx x

9 − x^2

3 cos θdθ 3 sin θ

9(1 − sin^2 θ)

dθ sin θ

csc θdθ =

ln | csc θ − cot θ| + C.

Using the fact that sin θ = x 3 , we find that csc θ = (^) x^3 and cot θ =

√ 9 −x 2 x. This yields the final answer:

1 3 ln

∣∣^3

x

9 − x^2 x

∣∣ + C.

m) U-substitution: Let u = 9 − x^2 , du = − 2 xdx. Then ∫ x^3 √ 9 − x^2

dx = −

9 − u √ u

du = − 9 u^1 /^2 +

u^3 /^2 + C = − 9

9 − x^2 +

(9 − x^2 )^3 /^2 + C.

n) Trig Substitution: Let x = 3 tan θ, dx = 3 sec^2 θdθ. Then ∫ dx √ 9 + x^2

3 sec^2 θdθ √ 9(1 + tan^2 θ)

sec θdθ = ln | sec θ + tan θ| + C

= ln

x^2 + 9 3

x 3

∣ +^ C.

o) Improper Integral! First, by definition, ∫ (^) ∞

1

xe−xdx = lim t→∞

∫ (^) t

1

xe−xdx.

c) Disks or Shells: V = 1283 π

d) Washers: V = π

0

(4 − (x − 4))^2 − (4 − (4 − x))^2 dx = 128π

  1. We break up the integral as follows:

4

(2f (x) − 5)dx = 2

4

f (x)dx −

4

5 dx = −49.

  1. The point of tangency is (0, g(0)) = (0, 0). The slope is g′(0) = esin 0^ = 1, since g′(x) = esin^ x^ by the FTC. Thus, the tangent line is y = x.
  2. a) arithmetic sequence, a = 3, d = 6; ak = 3 + 6(k − 1) = 6k − 3

b) method of finite differences; ak = k^2 − k + 4

c) arithmetic sequence; ak = 4 − 6(k − 1)

d) arithmetic denominators; ak = (^) 2+3(^1 k−1)

e) arithmetic numerator and denominator; ak = (^) 7+2(k+1k−1)

f) method of finite differences; ak = 12 k^3 − 32 k^2 + 2k + 1

  1. a) converges to 0

b) Since (^) xlim→∞

x ln(x−^1 ) = (^) xlim→∞ −lnx x = (^) xlim→∞ − 1 /x 1 = 0 by l’Hopital’s rule, then the sequence

must also converge to 0. Hence, lim k→∞

k ln(

k

c) Notice that lim k→∞

2 k+ 3 k^

= lim k→∞

)k = 0, since this is a geometric sequence with r = 23 < 1.

The sequence converges to 0.

d) Since (^) klim→∞

9 k^4 + 2k 2 k^2 + 4k + 7 = (^) klim→∞

k^2

9 + (^) k^23 k^2 (2 + (^) k^4 + (^) k^72 )

, the sequence converges to 32.

e) We’ll work instead with the continuous function f (x) =

x + 5 x

)x

. Let y =

x + 5 x

)x

. Then

ln y = (^) xlim→∞ x ln

x + 5 x

= (^) xlim→∞

ln

1 + (^5) x

1 x

Since the limit has 0/0 form, we can now apply l’Hopital’s rule, so the limit equals

xlim→∞

1 1+5/x ·^ (^

− 5 x^2 ) − 1 x^2

= (^) xlim→∞

1 + 5/x

So, (^) xlim→∞

x + 5 x

)x = e^5 and the corresponding sequence must also converge to e^5.

f) Write out the terms of the sequence: 0, 0, 0, ... Clearly, this sequence converges to 0.

  1. A. Converges to 6; geometric, a = 3, r = 1/ 2

B. Diverges; 4 x harmonic series

C. Diverges; geometric, r = 1, or use the nth^ term divergence test

D. Converges; comparison test Since 0 < (^3) k (^) +11^2 < (^32) k for k ≥ 2 and since

∑^ ∞

k=

3 k^ is a convergent geometric series (r = 1/3), the

series

∑^ ∞

k=

3 k^ + 11 must also converge by the comparison test for nonnegative series.

E. Diverges; geometric, r = e/ 2 > 1

F. Diverges Since limn→∞ 2 − (^) n^12 = 2 6 = 0, the series diverges by the nth^ term divergence test.

G. Converges to 1; telescoping series Notice that the terms break into partial fractions: (^) n(n^1 −1) = (^) n−^11 − (^1) n. Thus, the formula for the partial sum Sn simplifies to Sn = 1 − (^1) n and the series converges to limn→∞Sn = 1.

H. Converges; p-series with p = 4/ 3 > 1

I. Converges to − 45 ; geometric series combo The series is the difference of two convergent geometric series:

k= 1 4 k^ (with^ a^ = 1/4,^ r^ = 1/4) and

k=

3 k 4 k+1^ (with^ a^ = 3/16,^ r^ = 3/4). Thus, the series converges to^

1 3 −^

3 4 =^

− 5

J. Diverges; integral test We’ll consider the improper integral

2

4 x√ln x dx. Notice that ∫ (^) ∞

2

x

ln x

dx = lim t→∞

∫ (^) t

2

4(ln x)−^1 /^2 ·

x dx = lim t→∞ 8

ln t − 8

ln 2 = +∞,

so the improper integral diverges. So must the corresponding series by the integral test.

K. Converges; ratio test First, we simplify the term formula to

n=

5 n·(n+1) n!. Then we apply the ratio test:

nlim→∞

5 n+1·(n+2) (n+1)! 5 n·(n+1) n!

= (^) nlim→∞ 5(n + 2) (n + 1)(n + 1)

Since L = 0 < 1, the series converges.

L. Diverges; definition Consider the limit of the partial sums, which oscillate between 0, 1, and 2 and never approach a particular value. Since limn→∞Sn does not exist, the series diverges.

Another option: use the ratio test:

lim k→∞

∣∣^2

k+2 (^) · xk+ 2 k+1^ · xk

∣∣ = lim k→∞ 2 |x| = 2|x|,

so the series converges when 2|x| < 1 or − 1 / 2 < x < 1 /2. Checking the endpoints leads to the final answer.

B. Geometric power series with a = 1/5, r = 3 − 5 x This converges for |r| = | 3 − 5 x| < 1 or − 5 < 3 − x < 5 or − 8 < −x < 2 or − 2 < x < 8. Thus, the interval of convergence is (− 2 , 8). The sum is (^1) −(3^1 /−^5 x)/ 5 = (^5) −(3^1 −x) = (^) 2+^1 x.

C. This power series is the (term-by-term) antiderivative of ∑^ ∞

k=

(x + 1)k^ =

x + 1 2

(x + 1)^2 2

a geometric series that converges to (^1) −^1 (x/^2 +1) = − 2 x^1 when |x + 1| < 1, i.e. on the interval (− 2 , 0). Thus, the original series has the same radius of convergence, and its sum is an antiderivative of − 2 x^1.

Interval of convergence: We must check the endpoints. For x = −2,

k=

(−1)k 2 k is a multiple of the alternating harmonic series, so it converges. For x = 0, the power series becomes

k= 1 2 k , a divergent multiple of the harmonic series. Thus, the interval of convergence is [− 2 , 0).

Sum: We claimed above that

∑^ ∞

k=

(x + 1)k 2 k

ln |x| + C for some C. To determine C, we

evaluate both sides at the base point, x 0 = −1 and get C = 0. Thus, the sum of the series is − 12 ln |x|.

  1. A. Apply the ratio test:

lim j→∞

∣∣^ (3x)

j+1 (^) · j 3 (j + 1)^3 (3x)j

∣∣ = lim j→∞ | 3 x| · j^3 j^3 + 3j^2 + 3j + 1

= | 3 x|.

The series converges if L = | 3 x| < 1 or − 13 < x < 13. Now we test the endpoints.

For x = − 13 , the power series becomes

∑^ ∞

j=

(−1)j j^3 , which converges by the AST.

For x = 13 , the power series becomes

∑^ ∞

j=

j^3 , which converges by the p-test (p = 3).

So, the interval of convergence is [− 1 / 3 , 1 /3].

B. Apply the ratio test:

lim j→∞

∣∣^ (3x)

j+1 (^) · j! (j + 1)!(3x)j

∣∣ = lim j→∞ | 3 x| ·

j + 1

Since L = 0 < 1 for all x, the interval of convergence is (−∞, ∞).

C. The interval of convergence is [0, 8).

  1. p 3 (x) = 3 + 16 (x − 9) − 2161 (x − 9)^2 + 38881 (x − 9)^3. We can use p 3 (10) to approximate f (10) =

p 3 (10) = 3. 162294239... while the actual value of

10 is 3. 16227766 .... The error is less than .000017.

  1. f (x) = ln(1 − x) based at x 0 = 0

a) l(x) = −x

b) p 4 (x) = −x − x 2 2 −^

x^3 3 −^

x^4 4

c)

k=1 −^ xk k

d) Apply the ratio test: lim k→∞

xk+ k + 1

k xk^

= lim k→∞ |x| · k k + 1

= |x|. Thus, the series converges for

|x| < 1. We’ll check the endpoints.

For x = −1, we have

k=1 −^

(−1)k k , the alternating harmonic series, which converges. For x = 1, we have

k=1 −^ 1 k , the (negative) harmonic series, which diverges. The interval of convergence is [− 1 , 1).

B. f (x) = 1 − cos(2x) based at x 0 = π

a) l(x) = 0

b) p 4 (x) = 2(x − π)^2 − 23 (x − π)^4

c) S(x) =

(x − π)^2 −

(x − π)^4 +

(x − π)^6 −... =

∑^ ∞

k=

(−1)k+1^ · 22 k^ · (x − π)^2 k (2k)!

d) Same as the Taylor series for cos(x): (−∞, ∞)

C. f (x) = e−x^ based at x 0 = 0

a) l(x) = 1 − x

b) p 4 (x) = 1 − x + 12 x^2 − 16 x^3 + 241 x^4

c) S(x) =

k=

(−x)k k! d) Same as the Taylor series for ex: (∞, ∞)

  1. The second order Maclaurin polynomial is p 2 (x) = 0 + x + 0x^2 = x. Notice that p 2 (1) ≈ arctan(1), so arctan 1 ≈ 1. The error in this approximation is |arctan 1 − 1 | = |π/ 4 − 1 | < .215. At x = 2, the two functions are further apart: p 2 (2) = 2 is an approximation for arctan 2 with error less than .9.