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Final Exam Review Key–Math 132
- 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
- We break up the integral as follows:
4
(2f (x) − 5)dx = 2
4
f (x)dx −
4
5 dx = −49.
- 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.
- 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
- 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.
- 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|.
- 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).
- 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.
- 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: (∞, ∞)
- 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.
