Mathematics 1220
Final Exam Review Answers
Fall 2006
• Chapter 7: Transcendental Functions and First Order Differential Equations.
dy
2
= ln 7(2x + 1)7(x+1) −x
dx
x
dy
= ex sech2 xetanh(e )
(c)
dx
1. (a)
2. (a)
(b)
(c)
(d)
dy
− sin x
=√
= −1
dx
1 − cos2 x
dy
(d)
= xx (ln x + 1)
dx
Z
Z
1
u = sin x ⇒ du = cos xdx, cot x dx =
du = ln |u| + C = ln | sin x| + C
u
u = x2 − 1 ⇒ du = 2xdx,
2
Z
Z
10u
10x −1
1
u
x2 −1
10 du =
+C =
+C
x10
dx =
2
2 ln 10
2 ln 10
u
Z
Z = 3x ⇒ du = 3dx,
1
1
1
dx
=
du = 13 tan−1 u + C = 31 tan−1 3x + C
2
1 + 9x
3 1 + u2
2
u
Z = tanhx ⇒ du = sech
Z xdx,
sech2 x tanhx dx =
(b)
u du = 12 u2 + C = 21 tanh2 x + C
3. The doubling period implies 2 = e2k , so k = ln 2/2. You want to solve 1, 000, 000 =
100
10, 000e(ln 2/2)t for t = time in days. You find t = 2 ln
ln 2 , which is approximately
13 days, so you will definitely not be sick the day of the exam.
1
x
x , y = x(e + C)
2
2
ex , y = 12 + Ce−x
4. (a) Integrating factor:
(b) Integrating factor:
• Chapter 8: Techniques of Integration
1. Z
Trig techniques:
sin3/5 (2x + 1) cos3 (2x + 1) dx = 21 ( 85 sin8/5 (2x + 1) −
2. Trig techniques:
Z
cos4 x dx = 83 x +
5
16
5
18
sin18/5 (2x + 1)) + C
sin(2x) + C
3. Rationalizing substitution:
!
Z
3
2x − 1
−1 x − 3
p
dx = 6 ln
+ 5 tan
+C
x2 − 6x + 18
3
(x − 3)2 + 9
4. Z
Rationalizing substitution:
p
5 − 4x − x2 dx =
5. Integration by Parts:
6. Integration by Parts:
9
2
Z
Z
sin−1
x+2
3
+ 12 (x + 2) 9 − (x + 2)2 + C
p
ln(x2 + 4) dx = x ln(x2 + 4) − 2x + 4 tan−1
x cot2 x dx = x cot x − ln | sin x| − 12 x2 + C
x
2
+C
x+9
dx = ln x + 31 tan−1 x3 − 12 ln |x2 + 9| + C
3
x + 9x
Z
x4
8. Partial Fractions:
dx = − 31 x3 − x + 12 ln |1 − x| + 12 ln |1 + x| + C
1 − x2
7. Partial Fractions:
Z
Z
9. Integration by Parts:
ex/3 sin 3x dx =
3 x/3
sin 3x
82 e
−
27 x/3
cos 3x
82 e
+C
• Chapter 9: Improper Integrals
√
1. ln x < x1/4 and x3/2 < x3 + 2x + 1 imply
√
Therefore,
Z
∞
1
2.
Z
∞
0
ln x
x1/4
x1/4
1
<√
< 3/2 = 5/4 .
3
3
x
x
x + 2x + 1
x + 2x + 1
ln x
√
dx ≤
x3 + 2x + 1
Z
1
∞
x5/4
1
dx < ∞, and so it converges.
e−x cos x dx = e−x sin x − e−x cos x |∞
0 = 1, therefore converges.
3. For u = − ln(cos x),
Z
π/2
π/3
tan x
dx =
(ln cos x)2
Z
∞
ln 2
1
du < ∞, therefore converges.
u2
• Chapter 10: Power Series Representations of Functions
1. (a)
∞
X
−2n
e
=
n=0
∞ X
1 n
n=0
converges to
(b)
∞ X
1
k=1
lim
n→∞
1.
(c)
e2
1
1−e−2
1
√ −√
k+1
k
n
X
1
is the geometric series with r = e−2 < 1, therefore
= lim
n+1
X
n→∞
1
√ −
√
k k=2 k
k=1
!
n X
1
1
√ −√
k+1
k
k=1
= lim
n→∞
1
1− √
n+1
= lim
n→∞
n
X
1
n
X
1
√ −
√
k k=1 k + 1
k=1
= 1, therefore converges to
∞
∞
X
X
1
1
1
1
=
diverges, since
is the harmonic series.
16, 000k
16, 000 k=1 k
k
k=1
k=1
∞
X
∞ 1
1
dx =
du <
π
π
x(ln x)
2
ln 2 u
∞
X
1
∞, since π > 1. Therefore, by the integral test,
converges.
n(ln
n)π
n=2
(d) For the substitution u = ln x, the integral
(e) lim
1
n→∞ n2/3
Since
2
3
Z
∞
= 0, so by the alternating series test,
< 1, the series
converge absolutely.
Z
∞
X
(−1)n+1
n=1
∞
X
n=1
1
n2/3
1
n2/3
converges.
diverges, so the alternating series does not
2(n + 1)
2n+1 (n + 1)! (n + 2)!
= lim
= 2 implies
n
n→∞
n→∞
(n + 3)!
2 n!
(n + 3)
∞
X
2n n!
the series
diverges.
(n + 2)!
n=1
(f) By the ratio test, lim
!
=
(n + 1)!|x + 1|n+1
3n
(n + 1)|x + 1|
< 1 only when x =
= lim
n+1
n
n→∞
n→∞
3
n!|x + 1|
3
−1.
2. (a) lim
P
n3 + 1
|x|n+1
(−1)n
= |x| < 1, plus ∞
n=0 n3 +1 converges by the altern→∞ (n + 1)3 + 1 |x|n
P
1
nating series test and ∞
n=0 n3 +1 and converges by the limit comparison test
(with bn = n13 ), so the convergence set is the interval [−1, 1].
(b) lim
2|x|
2n+2 |x|n+1 (2n)!
= 0 < 1 for all x. There= lim
n+1
n
n→∞ (2n + 2)(2n + 1)
n→∞ (2n + 2)! 2
|x|
fore, the convergence set is (−∞, ∞)
(c) lim
3. (a) If g(x) =
1)xn
∞
∞
X
X
1
1
′
=
xn , then f (x) =
=
−g
(−x)
=
(−1)n+1 (n+
1 − x n=0
(1 + x)2
n=0
(b) If g(x) = tan
−1
x=
n=0
∞
X
(−1)n x2n+1
2n1
, then f (x) =
Z
x
0
tan−1 t
dt =
t
Z
x
0
g(t)
dt =
t
(2n + 1)2
n=0
(c) If g(x) = ex =
∞
X
xn
n!
n=0
4.
∞
X
(−1)n x2n+1
, then f (x) = xe−x = xg(−x) =
∞
X
(−1)n xn+1
n=0
n!
∞
∞
∞
X
X
X
1
1
=
(−1)n (x−1)n and 2 =
(−1)n+1 n(x−1)n−1 =
(−1)n (n+1)(x−1)n
x n=0
x
n=1
n=0
for all x in the interval (0, 2).
5. ln x =
Z
1
x
∞
X
(x − 1)n+1
1
dt =
(−1)n
.
t
n+1
n=0
• Chapter 11: Error in Numerical Techniques
1. |Rn (0.3)| ≤
(0.3)n+1
(n+1)!
approximation to cos x
2. |En | ≤
1
120n2
1
−3
(n+1)! ≤ 10
2
4
6
is 1− x2! + x4! − x6! ,
≤
so cos 0.3 is approximately 0.9553364875.
≤ 10−3 if n2 ≥ 9. Take n = 3, then
3. Let f (x) = sinx x . Then,
[0, 1]. Therefore, |En | ≤
Z
0.5
2
ex dx ≈ 0.54795
0
2 +x4 ) sin x+(−24x+4x3 ) cos x
(24−12x
≤ 51 for
f (4) (x) =
x5
1000
1
−5
4
if n ≥ 9 , so we may take n ≥ 4.
900n4 ≤Z 10
1 sin x
parabolic rule with n = 4 gives
0
4. n ≥ 40
when n ≥ 6. The sixth order Taylor
x
x in
The
dx ≈ 0.946087
5. We will approximate the root in the interval [2, 3] and take x1 = 1.2. Then,
x6 = 1.0893 is accurate to 3 decimal places.
6. Let x1 = 4.5. Then, x3 = 4.4934 approximates the fixed point accurately to 3
decimal places.
• Chapter 12: Conic Sections and Polar Coordinates
1. (a) Parabola, focus F = (1, 0), directrix x = −1
√
√
(b) Ellipse, foci F = (± 5, 0), directrices x = ± 9 5 5
√
√
(c) Hyperbola, foci F = (±5 5, 0), directrices x = ±4 5
2. (a)
(b)
(c)
2
(u−2)2
− v3 = 1, angle of rotation θ = π4 , center (2, 0), hyperbola.
4
u2
v2
π
4 + 12 = 1, angle of rotation θ = 4 , center (0, 0), ellipse.
(u+2)2
v2
+ 25
= 1, angle of rotation θ = π6 , center (−2, 0), ellipse.
5
• Chapter 18: Second Order Linear Differential Equations
1. (a) y = C1 ex +C2 e−x
2. (a) yp =
2
3
(b) y = C1 ex +C2 xex
cos x + 16 sin x
(b) yp = 14 xe2x
(c) y = C1 e−x cos x+C2 e−x sin x
(c) yp = 21 x2 − x + 1