Math 2260 Final Exam Practice Problem Solutions
1. Write down an integral which will compute the length of the part of y = ln x for 1 ≤ x ≤ 2. Don’t
worry about evaluating this integral.
Answer: Since
dy
dx
= x1 , the arc length integral is
Z
1
2
s
2
Z 2r
1
1
1+
dx =
1 + 2 dx
x
x
1
Z 2r 2
x +1
dx
=
x2
1
√
Z 2
1 + x2
=
dx
x
1
2. Let V (a) be the volume obtained by rotating the area between the x-axis and the graph of y =
from x = a to x = 1 around the y-axis. What happens as a goes to zero?
1
x3/2
Answer: Using the cylindrical shell method, V (a) is given by
1
Z
V (a) =
2πx
dx
x3/2
a
Z
1
1
1
dx
1/2
x
a
√ 1
= 2π 2 x a
√ = 2π 2 − 2 a
√
= 4π(1 − a).
= 2π
Therefore, as a → 0, the quantity V (a) → 4π.
√
3. What is the volume of the solid obtained by rotating the part of the graph of y = cos x sin x between
x = 0 and x = π/2 around the x-axis?
Answer: Using the disk method, the volume will be
Z
π/2
Z
2
√
π cos x sin x dx =
0
π/2
π cos2 x sin x dx.
0
Then, letting u = cos x, we have that du = − sin x, and so
Z
π/2
π cos2 x sin x dx = −π
π/2
Z
0
cos2 x(− sin x) dx
0
0
Z
u2 du
= −π
1
= −π
u3
3
0
= −π 0 −
=
1
π
.
3
1
1
3
4. Find the volume obtained by rotating the region between the x-axis, the y-axis, and the line x + y = 1
around the line x = −2.
Answer: I plan to use the shell method. Since the curve is y = 1 − x, the shell based at x has radius
x − (−2) = x + 2 and height 1 − x, so the volume is
Z 1
Z 1
−x2 − x + 2 dx
2π(x + 2)(1 − x) dx = 2π
0
0
1
−x3
x2
−
+ 2x
3
2
0
−1 1
− +2
= 2π
3
2
7
= 2π
6
7π
=
.
3
√
5. Calculate the surface area obtained by rotating y = x around the x-axis for 3/4 ≤ x ≤ 15/4.
= 2π
Answer: Since
dy
dx
=
1
√
,
2 x
Z
the surface area integral is
15/4
√
s
2π x 1 +
3/4
1
√
2 x
2
Z
15/4
dx = 2π
√
r
x
1+
3/4
Z
15/4
= 2π
r
x+
3/4
1
dx
4x
1
dx.
4
Let u = x + 41 . Then du = dx and the above integral is equal to
Z
2π
4
√
u du = 2π
1
2 3/2
u
3
4
= 2π
1
16 2
28π
−
.
=
3
3
3
6. (a) Find the area between y = 3 sin(πx) and y = 6x for x > 0.
6
5
4
3
2
1
0.2
0.4
0.6
0.8
1.0
Answer: First, we need to find the points where the curves intersect. The curves intersect when
3 sin(πx) = 6x,
meaning that
sin(πx) = 2x,
which happens when x = 0 or x = 1/2. Therefore, the area of the given region will be given by
the definite integral
1/2 Z 1/2
3
−3
3
3
−3 cos(πx)
2
(3 sin(πx) − 6x) dx =
− 3x
= 0−
−
−0 = − .
π
4
π
π
4
0
0
2
(b) What is the volume obtained by rotating this region around the x-axis?
Answer: My plan is to use the washer method. The outer radius of the washer at x is 3 sin(πx),
whereas the inner radius is 6x, so the area of the washer at x is
2
A(x) = πR2 − πr2 = π (3 sin(πx)) − π(6x)2 = 9π sin2 (πx) − 36πx2 .
Therefore, the volume of the solid is
1/2
Z
1/2
Z
9π sin2 (πx) − 36πx2 dx
A(x) dx =
0
0
Z
1/2
Z
2
sin (πx) dx − 36π
= 9π
1/2
x2 dx
0
0
3 1/2
1 − cos(2πx)
x
dx − 36π
2
3 0
0
1/2
x sin(2πx)
3π
= 9π
−
−
2
4π
2
0
1
3π
= 9π
−
4
2
9π 3π
=
−
4
2
3π
.
=
4
Z
1/2
= 9π
(c) What is the volume obtained by rotating around the y-axis?
Answer: Using the shell method, when the radius of the cylindrical shell is x − 0 = x, the height
is 3 sin(πx) − 6x, so the volume of the solid is
Z
1/2
Z
1/2
2πx (3 sin(πx) − 6x) dx = 2π
0
Z
3x sin(πx) dx − 2π
0
1/2
6x2 dx.
(1)
0
1/2
R 1/2
Certainly 2π 0 6x2 = 2π 2x3 0 = π2 , which takes care of the second term. To evaluate the
first term, I will integrate by parts using
u = 3x
dv = sin(πx) dx
du = 3 dx
v=
− cos(πx)
π
Then
Z
2π
0
1/2
!
1/2 Z 1/2
−3x cos(πx)
3 cos(πx)
3x sin(πx) dx = 2π
+
dx
π
π
0
0
1/2 !
3 sin(πx)
= 2π 0 +
π2
0
=
6
.
π
Therefore, from equation (1), we have that the volume of the solid is equal to
6
π
− .
π
2
3
7. What is the volume obtained by rotating the region between y =
around the x-axis?
√ 1
x2 +4
and the x-axis for 0 ≤ x ≤ 3
Answer: Using the disk method, the radius of a cross-sectional disk is equal to
the disk at x is equal to
2
1
π
A(x) = π √
= 2
.
x +4
x2 + 4
Therefore, the volume of the solid is equal to
Z 3
Z 3
π
A(x) dx =
dx.
2+4
x
0
0
√ 1
,
x2 +4
so the area of
To evaluate this integral, make the trig substitution x = 2 tan θ. Then dx = 2 sec2 θ dθ and the above
integral becomes
Z arctan(3/2)
Z arctan(3/2)
2π sec2 θ
2π sec2 θ dθ
=
dθ
4 sec2 θ
4 tan2 θ + 4
0
0
Z arctan(3/2)
π
dθ
=
2
0
h π iarctan(3/2)
=
x
2 0
π
= arctan(3/2).
2
8. What is the length of the part of y = x3/2 from x = 1 to x = 6?
√
dy
Answer: Since dx
= 32 x1/2 = 32 x, the arc length formula is
s
2
Z 6
Z 6r
3√
9
1+
x dx =
1 + x dx.
2
4
1
1
Then we can evaluate this integral by letting u = 1 + 94 x. This yields du =
integral becomes
29/2
Z
4 29/2 √
4 2 3/2
u du =
u
9 13/4
9 3
13/4
√
√ !
8 29 29 13 13
√ −
=
27
8
2 2
√
√
58 58 − 13 13
=
.
27
R
9. Integrate x4 ln(2x) dx.
Answer: Integrate by parts with
u = ln(2x)
du =
1
dx
x
dv = x4 dx
v=
x5
.
5
Then the above integral becomes
x5 ln(2x)
−
5
Z
x4
x5 ln(2x) x5
=
−
+ C.
5
5
25
4
9
4
dx and so the above
10. Integrate
R
x2 cos(3x) dx.
Answer: Integrate by parts with
u = x2
dv = cos(3x) dx
du = 2x dx
v=
sin(3x)
3
Then the above integral is equal to
Z
Z
x2 sin(3x)
x2 sin(3x) 2
2x sin(3x)
x sin(3x) dx.
−
dx =
−
3
3
3
3
To evaluate the second integral, integrate by parts again, with
u=x
dv = sin(3x) dx
du = dx
v=
− cos(3x)
.
3
Then the above becomes
Z
x2 sin(3x) 2 −x cos(3x)
cos(3x)
x2 sin(3x) 2x cos(3x) 2 sin(3x)
−
+
=
+
−
+ C.
3
3
3
3
3
9
27
11. Integrate
R
x+2
x2 +8x+15
dx.
Answer: I plan to use partial fractions:
x2
x+2
A
B
x+2
=
=
+
+ 8x + 15
(x + 5)(x + 3)
x+5 x+3
means that
x + 2 = A(x + 3) + B(x + 5)
x + 2 = (A + B)x + (3A + 5B),
so we have
1=A+B
2 = 3A + 5B
Therefore, A = 1 − B, so the second line becomes
2 = 3(1 − B) + 5B = 3 − 3B + 5B = 3 + 2B,
and hence
2B = −1
so it follows that B = −1/2 and A = 1 − B = 3/2.
Therefore,
Z
3/2
1/2
−
dx
x+5 x+3
3
1
= ln |x + 5| − ln |x + 3| + C.
2
2
x+2
dx =
x2 + 8x + 15
Z 5
12. Integrate
R√
1 − 4x2 dx.
Answer: Make the trig substitution x =
Z r
1
2
sin θ. Then dx =
1
2
cos θ dθ and the above integral becomes
Z
1
1 p
2 1
1 − 4 · sin θ cos θ dθ =
1 − sin2 θ cos θ dθ
4
2
2
Z
1 √ 2
=
cos θ cos θ dθ
2
Z
1
cos2 θ dθ
=
2
Z
1
1 + cos(2θ)
=
dθ
2
2
1 θ sin(2θ)
=
+
+C
2 2
4
1
sin(2θ)
=
θ+
+C
4
2
2 sin θ cos θ
1
θ+
+C
=
4
2
1
= (θ + sin θ cos θ) + C,
4
using the trig identity sin(2θ) = 2 sin θ cos θ. Now, since x = 21 sin θ, it follows that θ = arcsin(2x),
√
sin θ = 2x, and cos θ = 1 − 4x2 . Therefore, converting the above back into the variable x yields
√
p
1
arcsin(2x) x 1 − 4x2
2
+
+ C.
arcsin(2x) + 2x 1 − 4x + C =
4
4
2
13. Integrate
R
sin2 (ln(x))
x
dx.
Answer: Let u = ln x. Then du =
Z
14. Integrate
R
et
e2t +4
1
x
dx, so the above becomes
sin2 (u) du =
Z
1 − cos(2u)
u sin(2u)
du = −
+ C.
2
2
4
dt.
Answer: Let u = et . Then du = et dt and the above becomes
Z
du
.
2
u +4
Now, let u = 2 tan θ. Then du = 2 sec2 θ dθ, so the above becomes
Z
Z
2 sec2 θ dθ
1
θ
=
dθ = + C.
2
2
2
4 tan θ + 4
But now, since θ = arctan(u/2) and u = et , this is just
arctan(et /2)
+ C.
2
15. Evaluate
R∞
e
1
x(ln x)2
dx.
6
Answer: Let u = ln x. Then du =
1
x
dx and the above integral becomes
∞
Z
1
1
du = lim
b→∞
u2
Z
b
u−2 du
1
b
−1
b→∞
u 1
−1
+1
= lim
b→∞
b
= lim
= 1.
16. Evaluate
R∞
0
dx
(x+1)(x+4)
dx.
Answer: Using partial fractions
1
A
B
=
+
(x + 1)(x + 4)
x+1 x+4
implies that
1 = A(x + 4) + B(x + 1)
1 = (A + B)x + (4A + B)
so
0=A+B
1 = 4A + B
Therefore B = −A and so the second equation becomes
1 = 4A − A = 3A,
so A = 1/3 and hence B = −A = −1/3.
Thus
Z
0
∞
Z
b
dx
0 (x + 1)(x + 4)
Z b
1/3
1/3
= lim
−
dx
b→∞ 0
x+1 x+4
b
1
1
ln(x + 1) − ln(x + 4)
= lim
b→∞ 3
3
0
b
1
x+1
= lim
ln
b→∞ 3
x+4 0
1
b+1
1
1
= lim
ln
− ln
b→∞ 3
b+4
3
4
1
1
= ln(1) − ln(1/4)
3
3
− ln(1/4)
=
.
3
dx
= lim
(x + 1)(x + 4) b→∞
7
17. Solve the initial-value problem
dy
+ y = 2,
dx
y(0) = 1.
Answer: My goal is to separate variables. First, notice that I can write
dy
= 2 − y.
dx
Dividing by 2 − y, multiplying by dx, and integrating yields
Z
Z
dy
= dx.
2−y
But these integrals are easy, so we get
− ln(2 − y) = x + C,
or
ln(2 − y) = −x − C.
Now I need to solve for y, so exponentiate both sides:
2 − y = e−x−C ,
or
2 − y = Ae−x
where A = e−C .
Hence
y = 2 − Ae−x .
But then, since we know y(0) = 1, we have that
y(0) = 2 − Ae−0
1 = 2 − A,
so A = 1 and the solution to the initial-value problem is
y = 2 − e−x .
18. Solve the initial-value problem
dy
= 1 − y + x2 − yx2 ,
dx
y(0) = 0.
Answer: Notice that the right hand side can be re-written as
dy
= (1 + x2 )(1 − y).
dx
Therefore, we can separate variables and integrate:
Z
Z
dy
= (1 + x2 ) dx,
1−y
so
− ln(1 − y) = x +
8
x3
+ C.
3
Now we solve for y, so multiply by − = 1 and then exponentiate both sides:
1 − y = e−x−
x3
3
−C
= e−C e−x−x
3
/3
3
= Ae−x−x
/3
where A = e−C . Hence,
3
y = 1 − Ae−x−x
/3
.
Now we can use the initial condition to solve for A:
y(0) = 1 − Ae−0−0
3
/3
0 = 1 − A,
so A = 1 and therefore the solution to the initial-value problem is
3
y = 1 − e−x−x
/3
.
∞
19. Consider the sequence (an )n=1 , where
an =
ln(1 + 3/n)
.
sin(2/n)
Does the sequence converge? If so, to what? If not, explain why.
Answer: Taking the limit yields
ln(1 + 3/n)
n→∞ sin(2/n)
ln(1 + 3/x)
= lim
x→∞ sin(2/x)
lim an = lim
n→∞
=
1
−3
1+3/x x2
lim
x→∞ cos(2/x) −2
x2
= lim
x→∞
=
3
2 cos(2/x) 1 + x3
3
,
2
where I got from the second to the third lines by L’Hôpital’s Rule, so the sequence converges to 32 .
∞
20. Consider the sequence (an )n=1 , where
an =
n+1
n+2
n
.
Does the sequence converge? If so, to what? If not, explain why.
Answer: The goal is to evaluate limn→∞ ln(an ); then we’ll have that
lim an = elimn→∞ ln(an ) .
n→∞
9
Now,
n n+1
n→∞
n+2
n+1
= lim n ln
n→∞
n+2
x+1
ln x+2
= lim
1
lim ln(an ) = lim ln
n→∞
x→∞
= lim
x
ln(x + 1) − ln(x + 2)
1
x
x→∞
= lim
x→∞
1
x+1
−
1
x+2
−1
x2
2
−x
x2
+
x→∞ x + 1
x+2
2
−x (x + 2) + x2 (x + 1)
= lim
x→∞
(x + 1)(x + 2)
−x2
= lim
x→∞ (x + 1)(x + 2)
−x2
= lim 2
x→∞ x + 3x + 2
= −1,
= lim
where I went from the fourth to the fifth lines using L’Hôpital’s Rule.
Therefore,
lim an = e−1 =
n→∞
1
.
e
21. Which of the following series converge?
P∞ n3
(a)
n=1 3n
Answer: Using the Ratio Test:
(n+1)3
n+1
lim 3 n3
n→∞
3n
(n + 1)3 3n
n→∞ 3n+1 n3
= lim
1 n3
n→∞ 3 (n + 1)3
1
= .
3
= lim
(b)
Since
P∞
1
3
< 1, the Ratio Test says that this series converges.
n3
n=1 1+n4
Answer: I will do a limit comparison to the series
n3
1+n4
3
n→∞ n
n4
lim
P n3
n4 :
n4
= 1.
n→∞ 1 + n4
= lim
P n3
P1
Since the series
n4 =
n diverges and the above limit is finite and positive, the Limit Comparison Test implies that the given series also diverges.
10
(c)
n3
n=1 1+n5
P∞
Answer: Since 1 + n5 > n5 , it follows that
n3
1
n3
<
= 2
1 + n5
n5
n
for all n, and so
(d)
∞
X
∞
X
n3
1
<
.
5
1+n
n2
n=1
n=1
Since the series on the right hand side converges, so does the given series.
P∞ cos2 n
n=1 1+e−n
Answer: Notice that
cos2 n
cos2 n
= lim
,
−n
n→∞
n→∞ 1 + e
1
which does not exist. Therefore, since the terms are not going to zero, the nth term test implies
that the series diverges.
P ∞ en
lim
(e)
n=1 n!
Answer: Using the Ratio Test:
en+1
(n+1)!
lim
en
n→∞
n!
en+1 n!
n→∞ (n + 1)! en
e
= lim
n→∞ n + 1
= 0.
= lim
Since 0 < 1, the Ratio Test implies that the series converges.
22. If you put (−1)n into each of the series in the preceding problem, which ones converge absolutely,
which converge conditionally, and which diverge?
Answer:
(a) The answer to the previous problem shows that series
P
3
(−1)n 3nn converges absolutely.
3
n
(b) The series cannot converge absolutely, but since the terms 1+n
4 satisfy the hypotheses of the
P
3
n n
Alternating Series Test, the series (−1) 1+n4 does converge, so we conclude that it converges
conditionally.
P
n3
(c) The answer to the previous problem shows that the series (−1)n 1+n
5 converges absolutely.
P
cos2 n
(d) The terms still don’t go to zero, so the series (−1)n 1+e
−n diverges.
P
n
(e) The answer to the previous problem shows that the series (−1)n en! converges absolutely.
23. Does the series
∞
X
(−1)n
n(ln(n))2
n=2
converge absolutely, converge conditionally, or diverge?
Answer: To determine whether the series converges absolutely, or not, consider the series of absolutely
values
∞ ∞
X
(−1)n X
1
=
.
n(ln(n))2 n(ln(n))2
n=2
n=2
11
By the Integral Test, this series converges if and only if the following improper integral converges:
Z
∞
2
1
dx = lim
b→∞
x(ln(x))2
To evaluate this integral, let u = ln(x). Then du =
Z
ln(b)
lim
b→∞
ln(2)
1
x
Z
b
2
1
dx.
x(ln(x))2
dx and the above integral becomes
ln(b)
1
−1
du
=
lim
b→∞
u2
u ln(2)
−1
1
= lim
+
b→∞ ln(b)
ln(2)
1
=0+
ln(2)
1
=
.
ln(2)
Therefore the integral converges and so the Integral Test says that the series of absolute values converges, meaning that the given series converges absolutely.
24. Find the interval of convergence for each of the following power series.
P∞ (x+2)n
(a)
n=1
n
Answer: Using the Ratio Test on the series of absolute values,
(x+2)n+1 n+1 |x + 2|n+1
n
= lim
lim n
(x+2)
n→∞
n→∞
n
+
1
|x
+
2|n
n n
= lim |x + 2|
n→∞
n+1
= |x + 2|.
Therefore, the series definitely converges absolutely when |x + 2| < 1.
Now check the endpoints. When x + 2 = 1, the series becomes
∞
∞
X
X
1
1n
=
,
n
n
n=1
n=1
which diverges.
When x + 2 = −1, the series becomes
∞
X
(−1)n
,
n
n=1
which converges. Therefore, the power series converges for
−1 ≤ x + 2 < 1,
or
−3 ≤ x < −1,
which means the interval of convergence is [−3, −1).
12
(b)
(x−3)n
n=1 1+n2
P∞
Answer: Using the Ratio Test on the series of absolute values,
(x−3)n+1 1+(n+1)2 |x − 3|n+1 1 + n2
= lim
lim n
(x−3)
n→∞ 1 + (n + 1)2 |x − 3|n
n→∞
1+n2 1 + n2
n→∞
1 + (n + 1)2
n2 + 1
= lim |x − 3| 2
n→∞
n + 2n + 2
= |x − 3|.
= lim |x − 3|
Therefore, the series definitely converges absolutely when |x − 3| < 1.
Now check the endpoints. When x − 3 = 1, the series becomes
∞
X
∞
∞
X
X
1n
1
1
=
<
,
2
2
2
1
+
n
1
+
n
n
n=1
n=1
n=1
which converges, so the series converges when x − 3 = 1.
When x − 3 = −1, the series becomes
∞
X
(−1)n
,
1 + n2
n=1
which converges by the Alternating Series Test.
Therefore, the given power series converges for
−1 ≤ x − 3 ≤ 1,
or
2 ≤ x ≤ 4,
(c)
so the interval of convergence is [2, 4].
P∞ (−1)n (x−1)n
n=1
nen
Answer: Using the Ratio Test on the series of absolute values,
(−1)n+1 (x−1)n+1 (n+1)en+1 |x − 1|n+1
nen
= lim
lim n
n
n+1
(−1) (x−1) n→∞
n→∞ (n + 1)e
|x − 1|n
nen
|x − 1| n
e n+1
|x − 1|
=
.
e
= lim
n→∞
Therefore, the series definitely converges absolutely for |x−1|
< 1, meaning for |x − 1| < e.
e
Now, check the endpoints. When x − 1 = e, the series becomes
∞
∞
X
X
(−1)n en
(−1)n
,
=
nen
n
n=1
n=1
which converges.
13
When x − 1 = −e, the series becomes
∞
∞
∞
X
X
X
(−1)n (−1)n en
1
(−1)n (−e)n
=
=
,
n
n
ne
ne
n
n=1
n=1
n=1
which diverges.
Therefore, the given power series converges for
−e < x − 1 ≤ e,
or
1 − e < x ≤ 1 + e,
(d)
so the interval of convergence is (1 − e, 1 + e].
P∞
2 n
n=1 n x
Answer: Using the Ratio Test on the series of absolute values,
(n + 1)2 xn+1 (n + 1)2 |x|n+1
=
lim
lim
n→∞
n→∞
|n2 xn |
n2
|x|n
2
(n + 1)
= lim
|x|
n→∞
n2
= |x|,
so the series definitely converges absolutely for |x| < 1.
Now check the endpoints. When x = 1, the series becomes
∞
X
n2 1n =
n=1
∞
X
n2 ,
n=1
which diverges since the terms are going to infinity.
Likewise, when x = −1, the series becomes
∞
X
n2 (−1)n ,
n=1
which diverges since the terms are getting large in absolute value and hence not going to zero.
Therefore, the series converges for
−1 < x < 1,
so the interval of convergence is (−1, 1).
25. Find the Taylor series for
2
(a) e−x centered at x = 0
Answer: The Taylor series for ex centered at x = 0 is
ex = 1 + x +
∞
X
x2
x3
xn
+
+ ... =
.
2!
3!
n!
n=0
2
Then, substituting −x2 for x in the above yields the Taylor series for e−x centered at x = 0:
2
(−x2 )2
(−x2 )3
+
+ ...
2!
3!
4
6
x
x
= 1 − x2 +
−
+ ...
2!
3!
∞
X
(−1)n x2n
=
.
n!
n=0
e−x = 1 + (−x2 ) +
14
(b)
√
x centered at x = 1
Answer: Remember that the Taylor series for a function f (x) centered at x = c is given by
f (x) =
∞
X
f (n) (c)
(x − c)n .
n!
n=0
Therefore, we need to determine the values of the derivatives of f (x) =
compute the derivatives of f (x):
√
x at x = 1. First,
1
f 0 (x) = √
2 x
−1
f 00 (x) = 3/2
4x
3
000
f (x) = 5/2
8x
−15
(4)
f (x) =
16x7/2
..
.
Therefore, values of f and its derivatives at x = 1 are:
√
f (1) = 1 = 1
1
1
f 0 (1) = √ =
2
2 1
−1
−1
f 00 (1) =
=
4
4 · 13/2
3
3
000
=
f (1) =
8
8 · 15/2
−15
−15
(4)
f (1) =
=
16
16 · 17/2
..
.
Therefore, the Taylor series for f is
1
1
3
15
f (x) = 1 + (x − 1) −
(x − 1)2 +
(x − 2)3 −
(x − 1)4 + . . .
2
4 · 2!
8 · 3!
16 · 4!
26. Estimate e−0.2 to the nearest 0.001. Explain your answer.
Answer: Since the Taylor series centered at x = 0 for ex is
ex = 1 + x +
x2
x3
+
+ ...,
2!
3!
and this series converges absolutely for all real numbers x, it follows that
e−0.2 = 1 + (−0.2) +
(−0.2)2
(−0.2)3
0.04 0.008
+
+ . . . = 1 − 0.2 +
−
+ ....
2!
3!
2
6
4
Since the series alternates and the next term in the series is (0.2)
= 0.0016
< 0.001, it follows that the
4!
24
approximation
0.04 0.008
e−0.2 ≈ 1 − 0.2 +
−
= 0.8 + 0.02 − 0.0013 ≈ 0.819
2
6
is correct to three decimal places.
15
27. Does the series
∞
X
n7 6n
n!
n=1
converge or diverge? Explain your answer.
Answer: Use the Ratio Test:
(n+1)7 6n+1
(n+1)!
lim
n7 6n
n→∞
n!
6
(n + 1)7 6n+1
n!
(n + 1)7
=0
· 7 n = lim
·
n→∞ n + 1
n→∞
(n + 1)!
n 6
n7
= lim
since the first term goes to zero and the second term goes to 1.
Therefore, since 0 < 1, the Ratio Test tells us that the series converges.
28. Does the series
∞
X
1
√
n+ n
n=2
converge or diverge? Explain your answer.
Answer:
P 1 For large n, the n should dominate the
series
n.
1√
n+ n
lim
1
n→∞
n
Therefore, since
P
√
n, so let’s do a limit comparison to the divergent
n
1
1
n
√ · = lim
√ = lim
= 1.
n→∞ n +
n+ n 1
n n→∞ 1 + √1n
P 1
√ .
diverges, so does
n+ n
1
n
= lim
n→∞
29. Find the limit
lim
x→0
cos x − 1
.
ex2 − 1
Answer: Recall that the Taylor series centered at x = 0 for cos(x) is
cos(x) = 1 −
Since ex = 1 + x +
x2
2!
+
x3
3!
x4
x6
x2
+
−
+ ....
2!
4!
6!
2
+ . . ., the Taylor series centered at x = 0 for ex is
2
ex = 1 + (x2 ) +
(x2 )3
x4
x6
(x2 )2
+
+ . . . = 1 + x2 +
+
+ ....
2!
3!
2!
3!
Therefore,
lim
x→0
x2
2!
1−
+
cos(x) − 1
= lim
x→0 1 + x2 +
ex2 − 1
2
= lim
x→0
= lim
x→0
= lim
x→0
1
=− .
2
16
x4
4!
x4
2!
− ... − 1
+ ... − 1
x4
24 − . . .
x4
2 + ...
− x2 +
x2 +
x2 − 12 +
x2 1 +
− 12 +
1+
x4
24
x2
2
− ...
+ ...
x2
24
x2
2
− ...
+ ...
30. Give a geometric description of the set of points (x, y, z) satisfying the following inequalities.
(a) x2 + y 2 + z 2 ≤ 1 and z ≥ 1/2.
Answer: Since x2 + y 2 + z 2 = 1 describes the surface of the sphere of radius 1, the points
satisfying the inequality x2 + y 2 + z 2 ≤ 1 are those points lying on the surface of this sphere or
inside it.
On the other hand, the plane z = 1/2 is the plane parallel to the xy-plane at height 1/2; the
points satisfying z ≥ 1/2 are those points lying in this plane or above it.
Therefore, the points satisfying both of the given inequalities are those points lying on or in the
sphere of radius 1 centered at the origin and above the plane at height 1/2; in other words, this
is the top one-fourth of the sphere of radius 1.
(b) |x| ≤ 1, |y| ≤ 1, and |z| ≤ 1.
Answer: If |x| ≤ 1, then −1 ≤ x ≤ 1, so points satisfying this inequality are those points lying
no further than a distance 1 from the yz-plane, and similarly for the other two inequalities.
Therefore, the points satisfying all of the above inequalities are those points lying within the cube
of side length 2 which is centered at the origin.
31. Determine the angle between the vectors
*
*
*
*
u = ı + 3  + 2k
*
*
*
*
v = −2 ı +  − 2 k
and
Answer: Recall that
*
*
*
*
u · v = | u|| v| cos θ,
*
*
where θ is the angle between u and v, so it follows that
*
cos θ =
*
u·v
* * .
| u|| v|
Now, by definition of the dot product,
*
*
u · v = (1)(−2) + (3)(1) + (2)(−2) = −2 + 3 − 4 = −3.
Also,
*
| u| =
p
12 + 32 + 22 =
√
and
*
| v| =
p
(−2)2 + 12 + (−2)2 =
1+9+4=
√
√
4+1+4=
14
√
9 = 3.
Therefore,
*
cos θ =
*
*
−3
u·v
−1
√ =√ ,
* * =
| u|| v|
3 14
14
*
so the angle between u and v is
θ = arccos
−1
√
14
.
(If you’re curious, this turns out to be about 105◦ .)
32. Write the vector
*
*
*
*
u = ı +  + 5k
*
*
*
*
*
as the sum of two vectors, one parallel to v = −2 ı −  − k and one perpendicular to v.
17
*
*
*
*
Answer: To get the component of u parallel to v, take the projection of u onto v:
*
*
proj*
vu =
*
u · v*
* v.
v·v
*
Now,
*
*
u · v = (1)(−2) + (1)(−1) + (5)(−1) = −2 − 1 − 5 = −8
and
*
*
v · v = (−2)2 + (−1)2 + (−1)2 = 4 + 1 + 1 = 6,
so
*
*
proj*
vu =
*
−8 * * * 8 * 4 * 4 *
u · v*
−2 ı −  − k = ı +  + k.
*
* v =
6
3
3
3
v·v
*
*
Therefore, the component of u perpendicular to v is just
*
8* 4* 4*
5 * 1 * 11 *
*
*
*
*
u − proj*
ı +  + k =− ı −  +
k.
v u = ( ı +  + 5 k) −
3
3
3
3
3
3
*
*
*
Hence, u can be written as the sum of a vector parallel to v and a vector perpendicular to v as follows:
8* 4* 4*
5 * 1 * 11 *
*
*
*
*
*
u = proj*
ı +  + k + − ı −  +
k
v u + ( u − proj v u) =
3
3
3
3
3
3
33. Find the area of the parallelogram spanned by the vectors
*
*
*
*
u = −ı +  − k
and
*
*
*
*
v = 3 ı + 2  + k.
*
*
Answer: Remember that the area of the parallelogram spanned by u and v is equal to the magnitude
*
*
of the cross product of u and v.
Hence, to solve this problem, I should first compute the cross product:
* * * ı

k *
*
u × v = −1 1 −1 3
2
1 *
*
*
= ı ((1)(1) − (−1)(2)) −  ((−1)(1) − (−1)(3)) + k((−1)(2) − (1)(3))
*
*
*
= 3 ı − 2  − 5k
Now, the area of the parallelogram is equal to
p
√
√
*
*
| u × v| = 32 + (−2)2 + (−5)2 = 9 + 4 + 25 = 38.
18