MATH1120 Calculus II
Solutions to Selected Review Problems II
November 3, 2013
Z
(1) Evaluate
cos(ln x)dx.
Solution: We apply integration by parts. Let u = cos(ln x) and dv = dx. Then
v can be taken to be x.
Z
Z
cos(ln x)dx = x cos(ln x) − xd(cos(ln x))
Z 1
dx
= x cos(ln x) − x − sin(ln x) ·
x
Z
= x cos(ln x) + sin(ln x)dx
Z
= x cos(ln x) + x sin(ln x) − xd (sin(ln x)) (apply integration by parts once more)
Z
1
= x cos(ln x) + x sin(ln x) − x cos(ln x) · dx
x
Z
= x cos(ln x) + x sin(ln x) − cos(ln x)dx
Z
2 cos(ln x)dx = x cos(ln x) + x sin(ln x) + C1
Z
x cos(ln x) + x sin(ln x)
C1
cos(ln x)dx =
+ C (C =
)
2
2
Z
2
xn−1 ex
n−1
2
(2) Let In = xn ex dx, n ≥ 2. Show that In =
−
In−2 .
2
2
2
ex
2
. IntegraSolution: Let u = xn−1 and dv = xex dx. So v can be taken to be
2
tion by parts gives
Z x2
2
xn−1 ex
e
In =
−
d(xn−1 )
2
2
Z
2
2
xn−1 ex
(n − 1)xn−2 ex
=
−
dx
2
2
2
=
xn−1 ex
n−1
−
In−2
2
2
Remark 0.1. Reduction formulas are usually proved using integration by parts. In
this question one can guess the right choice of u and dv by inspecting the boundary
2
xn−1 ex
.
term (i.e. the first term of the RHS)
2
1
2
Z
−1 √
−1 √
Z
1
x
√
dx
xdx = x sin
x−
(3) (a) Show that sin
2
x − x2
√
Solution: Let u = sin−1 x and dv = dx. Integration by parts gives
Z
Z
√
−1 √
−1 √
sin
xdx = x sin
x − xd(sin−1 x)
Z
1
1
−1 √
= x sin
x− x· √
· √ dx
1−x 2 x
Z
√
1
x
√
dx
= x sin−1 x −
2
x − x2
Z
x
(b) Evaluate √
dx. (Hint: Complete squares for x−x2 and use a trigonom2
x−x
etry substitution)
Solution:
Z
Z
x
x
√
q dx =
2 dx
2
x−x
1 2
− x− 1
2
2
1
1
π
Let x − = sin θ. Since we are dealing with an indefinite integral, − ≤
2
2
2
π
θ ≤ by convention. Then dx = 21 cos θdθ.
2
Z
Z 1
sin θ + 12 21 cos θdθ
x
2
q 1
dx =
1 2
1 2
2 | cos θ|
−
x
−
2
2
1
Z 1
1
π
π
2 sin θ + 2
2 cos θdθ
(Since − ≤ θ ≤ , cos θ ≥ 0)
=
1
2
2
2 cos θ
Z 1
1
=
sin θ +
dθ
2
2
1
1
= − cos θ + θ + C
2
2
1p
1
=−
1 − (2x − 1)2 + sin−1 (2x − 1) + C
2
2
p
1
−1
= − x − x2 + sin (2x − 1) + C
2
Z
(4) (a) Evaluate sin5 3x cos2 3xdx.
Solution: Let u = 3x. Then du = 3dx.
Z
Z
du
sin5 3x cos2 3xdx = sin5 u cos2 u ·
3
Z
1
=
sin5 u cos2 udu
3
3
Since the power of sin u is odd, we make another substitution v = cos u. Then
dv = − sin udu.
Z
Z
1
1
sin5 u cos2 udu =
− sin4 u cos2 u(− sin u)du
3
3
Z
1
−(1 − cos2 u)2 cos2 u(− sin u)du
=
3
Z
1
=
−(1 − v 2 )2 v 2 dv
3
Z
1
=−
(v 2 − 2v 4 + v 6 )dv
3
1 v 3 2v 5 v 7
=−
+C
−
+
3 3
5
7
cos3 3x 2 cos5 3x cos7 3x
=−
+
−
+C
9
15
21
Z √ 2x
e −4
(b) Evaluate
dx.
ex
Solution: Let ex = 2 sec θ. Since ex is always positive, so is 2 sec θ. Therefore
π
2 sec θ tan θdθ
0 ≤ θ < . It follows that ex dx = 2 sec θ tan θdθ =⇒ dx =
=
2
ex
2 sec θ tan θdθ
= tan θdθ.
2 sec θ
Z √ 2x
Z
e −4
2 tan θ
=
· tan θdθ
x
e
2 sec θ
Z
sec2 θ − 1
dθ
=
sec θ
Z
= (sec θ − cos θ)dθ
= ln | tan θ + sec θ| − sin θ + C
π
= ln(tan θ + sec θ) − sin θ + C (0 ≤ θ < )
2
! √
√
x
2x
2x
e −1 e
e −4
= ln
+
+C
−
2
2
ex
(5) (a) Use Heaviside’s method to resolve
250k 2 + 400k + 100
(5k + 1)(5k + 2)(5k + 6)(5k + 7)
into partial fractions. Note that Heaviside’s method can be applied to proper
fractions only. If the fraction is improper, then do long division first.
4
Solution:
A
B
C
D
250k 2 + 400k + 100
=
+
+
+
(5k + 1)(5k + 2)(5k + 6)(5k + 7)
5k + 1 5k + 2 5k + 6 5k + 7
2
250 − 15 + 400 − 15 + 100
=1
A=
5 − 15 + 2 5 − 15 + 6 5 − 15 + 7
Similarly, we have B = 1, C = D = −1. So
1
1
1
1
250k 2 + 400k + 100
=
+
−
−
(5k + 1)(5k + 2)(5k + 6)(5k + 7)
5k + 1 5k + 2 5k + 6 5k + 7
∞
X
250k 2 + 400k + 100
(5k + 1)(5k + 2)(5k + 6)(5k + 7)
k=1
Solution: By part (a), the n-th partial sum of the series is
(b) Use part (a) and method of telescoping to find
sn =
1
1
1
1
13
1
1
+
−
−
=
−
−
5(1) + 1 5(1) + 2 5n + 6 5n + 7
42 5n + 6 5n + 7
13
So the sum is lim sn = .
n→∞
42
Z
dx
. (Hint: x4 + 4 = (x2 + 2x + 2)(x2 − 2x + 2))
(c) Evaluate
x4 + 4
1
Solution: We first resolve 4
into partial fractions. Note that x4 + 4 =
x +4
1
(x2 + 2x + 2)(x2 − 2x + 2), where the two factors are irreducible. Let 4
=
x +4
Ax + B
Cx + D
+ 2
. Clearing denominators, we have
2
x + 2x + 2 x − 2x + 2
1 = (Ax + B)(x2 − 2x + 2) + (Cx + D)(x2 + 2x + 2)
1 = (A + C)x3 + (−2A + B + 2C + D)x2 + (2A − 2B + 2C + 2D)x + (2B + 2D)
5


A+C
=0
A = 81






−2A + B + 2C + D = 0
B = 1
4 .
=⇒


C = − 18
2A − 2B + 2C + 2D = 0






D = 14
2B + 2D
=1
Z
Z
Z
Z
1
1
1
1
− 18 x + 14
− 81 x + 41
8x + 4
8x + 4
dx
+
dx
=
dx
+
dx
x2 + 2x + 2
x2 − 2x + 2
(x + 1)2 + 1
(x − 1)2 + 1
Z
Z 1
1
− 18 tan α + 18
2
8 tan θ + 8
·
sec
θdθ
+
· sec2 αdα
=
sec2 θ
sec2 α
(Let x + 1 = sec θ for the first integral and x − 1 = sec α
for the second one)
Z Z 1
1
1
1
=
dθ +
− tan α +
dα
tan θ +
8
8
8
8
1
1
1
1
= ln | sec θ| + θ − ln | sec α| + α + C
8
8
8
8
1
x2 + 2x + 2 1
=
ln
+ (tan−1 (x + 1) + tan−1 (x − 1)) + C
16 x2 − 2x + 2 8
√
2x
(d) Let R be the region bounded by the x-axis, the graph of p
(x2 + 1)(x2 − 2x + 1)
and the two vertical lines x = 0 and x = 4. Find the volume of the solid formed
by revolving R around
Z 4 the x-axis.2
Z 4
2x
2x2
Solution 1 : V = π
dx
=
π
. We
2
2
2
2
0 (x + 1)(x − 2x + 1)
0 (x + 1)(x − 1)
2x2
into partial fractions. Let
shall resolve the integrand 2
(x + 1)(x − 1)2
2x2
Ax + B
C
D
= 2
+
+
2
2
(x + 1)(x − 1)
x +1
x − 1 (x − 1)2
Clearing denominators, we have
2x2 = (Ax + B)(x − 1)2 + C(x2 + 1)(x − 1) + D(x2 + 1)
2x2 = (A + C)x3 + (−2A + B − C + D)x2 + (A − 2B + C)x + B − C + D


A+C
=0
A = −1






B = 0
−2A + B − C + D = 2
=⇒
. Hence


C =1
A − 2B + C
=0






B−C +D
=0
D =1
x
1
1
2x2
=− 2
+
+
2
2
(x + 1)(x − 1)
x + 1 x − 1 (x − 1)2
6
Note that the integrand is improper of type II because it has an infinite discontinuity
at x = 1. We should evaluate the volume integral by splitting the interval of
integration into two.
Z
4
π
0
2x2
dx = π
(x2 + 1)(x − 1)2
Z
1
0
Z 4
x
1
1
− 2
+
+
x + 1 x − 1 (x − 1)2
x
1
1
+
+
x2 + 1 x − 1 (x − 1)2
−
+π
1
dx
dx
Z 1
1
1
x
x
− 2
− 2
Consider the first term π
dx. Note that
+
+
dx
2
x + 1 x − 1 (x − 1)
x +1
0
0
Z 1
1
1
dx. But
is finite because it is not improper. It remains to find
+
(x − 1)2
0 x−1
Z
Z
1
0
1
1
1
+
dx = lim
x − 1 (x − 1)2
b→1−
Z
b
1
1
+
dx
(x − 1)2
0 x−1
1
= lim ln |b − 1| −
−1
b−1
b→1−
= DNE
So the first integral diverges As a result, the volume integral diverges, and the
volume is infinite.
Solution 2 : The volume integral is of type II because the integrand has an infinite
discontinuity at x = 1. So
Z
V =π
0
Z
Consider
1
4
1
2x2
dx +
(x2 + 1)(x − 1)2
Z
1
4
2x2
dx
(x2 + 1)(x − 1)2
2x2
dx. By letting u = x − 1, we have
(x2 + 1)(x − 1)2
Z
1
4
2x2
dx =
2
(x + 1)(x − 1)2
Z
3
2(1 + u)2
du
((1 + u)2 + 1)u2
3
2u2 + 4u + 2
du
u4 + 2u3 + 2u2
0
Z
=
0
2u2 + 4u + 2
. A comparison function g(u) can be obtained in this
u4 + 2u3 + 2u2
type II case by picking the lowest order terms of both the numerator and denomi2
1
nator. So g(u) = 2 = 2 . Since both f (u) and g(u) are positive and continuous
2u
u
Let f (u) =
7
for u ∈ (0, 3],
lim
u→0+
Z
2u2 + 4u + 2
f (u)
= lim 2
g(u) u→0+ u + 2u + 2
=1
6= 0
3
1
du diverges, we can apply the Limit Comparison Test and conclude that
2
0 u
2x2
dx diverges, and so does the original volume integral.
2
2
1 (x + 1)(x − 1)
Z ∞
1
(6) (a) Determine the convergence of the improper integral
dx by
x
−x
−∞ e + e
(i) evaluating it directly.
du
Solution: Let u = ex , then du = ex dx and dx =
.
u
Z ∞
Z ∞
Z 0
1
1
1
dx
=
dx
+
dx
x
−x
x
−x
x
−x
e +e
−∞ e + e
0
−∞ e + e
Z b
Z 0
1
1
= lim
dx
+
lim
dx
x
−x
x
a→−∞ a e + e−x
b→∞ 0 e + e
Z eb
Z 1
1
du
du
1
= lim
·
+ lim
·
a→−∞ ea u + u−1
b→∞ 1 u + u−1
u
u
Z eb
Z 1
1
1
du + lim
du
= lim
2
2
a→−∞
b→∞ 1 1 + u
ea u + 1
and
Z 4
b
= lim [tan−1 u]e1 + lim [tan−1 u]1ea
b→∞
a→−∞
π
=
2
So the improper integral converges.
(ii) using one of the comparison tests.
Solution:
Z ∞
Z 0
Z ∞
1
1
1
dx =
dx +
dx
x
−x
x
−x
x
−x
e +e
0
−∞ e + e
−∞ e + e
We should determine the convergence of the two integrals on the RHS
separately. Note that as x → ∞,Z ex is the dominant term of ex + e−x .
∞
1
1
−x
Since x
<
=
e
and
e−x dx converges, the first integral
e + e−x
ex
0
also converges by the Direct Comparison Test. Similarly, as x → −∞,
1
1
e−x is the dominant term of ex + e−x . Since x
< −x = ex and
−x
e +e
e
8
Z
0
ex dx converges, the second integral converges as well by the Direct
−∞
Comparison
Test. All in all, the original improper integral converges.
Z ∞
1
(b) Show that
1 dt is divergent.
1
(t3 + 1) 3
1
1
Solution: Let f (t) =
. Both are continuous and positive
1 and g(t) =
t
(t3 + 1) 3
on [1, ∞). Note that
f (t)
t
1
= lim
1 = lim
1 = 1
t→∞ g(t)
t→∞ (t3 + 1) 3
t→∞
1 + t13 3
Z ∞
g(t)dt = lim [ln t]b1 = ∞. So by Limit Comparison Test, the
and that
lim
1
b→∞
given integral is divergent.
Remark 0.2. Since t3 is the dominant term in the polynomial t3 + 1 when
t → ∞, t3 + 1 and t3 have comparable growth rates. Therefore we choose
1
1
g(t) =
.
1 =
3
t
(t ) 3
(c) Find
Rx
lim
x→∞
1
1 (t3 +1) 13
dt
x
(Hint: Use L’Hopital’s rule and FTC. Make sure you explain why you can
apply L’Hopital’s rule)
Solution: Since the numerator is divergent by part (a), the fraction is of the
∞
form
. Thus L’Hopital’s rule can be applied.
∞
Rx
Rn
1
1
1 (t3 +1) 13 dt
1 (t3 +1) 13 dt
lim
= lim
x→∞
n→∞
n
R xx 1
d
1 dt
dx 1
(t3 +1) 3
= lim
(by L’Hopital’s Rule)
x→∞
1
1
= lim
(By FTC)
x→∞ (x3 + 1) 13
=0
(7) Find the following limits
1
(a) lim (3n + 5n ) n
n→∞
9
Solution:
n
1
n
3
n
+1
lim (3 + 5 ) = lim 5
n→∞
n→∞
5
n
1
n
3
= lim 5
+1
n→∞
5
n
limn→∞ 1
n
3
= 5 lim
+1
n→∞ 5
n
3
= 0 by the commonly occurring limit lim xn = 0 for |x| < 1)
= 5 ( lim
n→∞
n→∞ 5
n
n
1
n
1
1 ln n
(b) lim
n→∞ n
Solution:
1
1
1
1 ln n
= lim e ln n ln( n )
lim
n→∞
n→∞ n
= lim e
− ln n
ln n
n→∞
= e−1
1
2 + sin n
when n is odd and an =
when n is even.
n
n2
2 + sin n
2+1
3
Solution: When n is even, an =
≤
= . When n is odd,
n2
n
n
1
3
an = < . So for any n,
n
n
(c) lim an , where an =
n→∞
0 ≤ an ≤
3
n
3
Since lim
= 0, lim an = 0 by Sandwich Theorem.
n→∞ n n n→∞
1
(d) lim 1 − 2
n→∞
n
Solution:
2 ! n1
1 n
1 n
lim 1 − 2
= lim
1− 2
n→∞
n→∞
n
n
1
= lim (e−1 ) n
n→∞
=1
(by the commonly occurring limit lim
n→∞
1
n
(by the commonly occurring limit lim x = 1)
n→∞
1+
x n
= ex )
n
10
Alternative solution 1 :
1 n
1 n
1 n
= lim 1 +
1−
lim 1 − 2
n→∞
n→∞
n
n
n
n
1 n
1
lim 1 −
= lim 1 +
n→∞
n→∞
n
n
= e1 · e−1
=1
Alternative solution 2 :
1 n
n ln 1− 12
n
lim 1 − 2
= lim e
n→∞
n→∞
n
ln(1− 12 )
n
limn→∞
1
n
=e
limx→∞
ln 1− 12
x
1
x
limx→∞
x2 · 2
x2 −1 x3
− 12
x
=e
=e
limx→∞
=e
(
)
( )
−2x
x2 −1
(by L’Hopital’s Rule)
= e0
=1
p
p
(e) lim ( n2 + n − n2 − n)
n→∞
Solution:
√
√
p
p
p
p
n2 + n + n2 − n
2
2
2
2
√
lim ( n + n − n − n) = lim ( n + n − n − n) √
n→∞
n→∞
n2 + n + n2 − n
2n
√
= lim √
n→∞
n2 + n + n2 − n
2
q
= lim q
n→∞
1 + n1 + 1 − n1
=1
(8) Determine the convergence of the following series.
∞
X
1
p
(a)
n(n + 1)(n + 2)
n=1
Solution: Note that an > 0 for all n. Also
1
1
1
an = p
<√ = 3
3
n(n + 1)(n + 2)
n
n2
11
∞
X
1
∞
X
3
> 1. By the Direct Comparison Test,
an
3 converges because p =
2
n=1 n 2
n=1
converges. It converges absolutely because an > 0 for all n and |an | = an .
∞
X
(−1)n (n2 + 1)
(b)
2n2 + n − 1
n=1
n2 + 1
. Note that
2n2 + n − 1
1
lim un =
6= 0. So lim (−1)n un does not exist. By the n-term test, the
n→∞
n→∞
2
given series diverges.
∞
X
2n (n!)2
(c)
(2n)!
n=1
Solution: Note that each term of the given series is positive.
Solution: This series is alternating, where un =
2n+1 ((n+1)!)2
(2(n+1))!
lim
2n (n!)2
n→∞
(2n)!
2(n + 1)2
n→∞ (2n + 1)(2n + 2)
= lim
1
2
<1
=
By the Ratio Test, the given series converges.
∞
X
√
√
(d)
( n + 1 − n)
n=1
Solution: The series
is a telescoping
sum. Note that the n-th partial sum, sn ,
√
√
√
of the series is n + 1 − 1 = n + 1 − 1. The sum is by definition
lim sn = lim
n→∞
(e)
n→∞
√
n+1−1=∞
So the given series diverges.
∞
X
ln n
n=1
n
∞
X
ln n
1
and bn = , For n ≥ 3, an ≥ bn ≥ 0, and
bn
n
n
n=1
diverges. By the Direct Comparison Test, the given series diverges as well.
ln n
ln x
1 − ln x
Solution 2 : Let an =
. Let f (x) =
. Since f 0 (x) =
< 0
n
x
x2
for x ≥ 3, f (x) is decreasing for x ≥ 3. So similarly, for n ≥ 3, an ≥ an+1 .
Besides, an are positive. So the conditions for Integral Test are satisfied. Note
Solution 1 : Let an =
12
that
Z
3
∞
ln x
dx = lim
b→∞
x
Z
3
Z
b
ln x
dx
x
ln b
du
dx
(Let u = ln x. Then du =
)
b→∞ ln 3 u
x
= lim (ln(ln b) − ln(ln 3))
= lim
b→∞
=∞
So by Integral Test, the given series diverges.
ln n
(f) Let an = 2 . Note that for n sufficiently large, ln n < nc , where c is some
n
positive constant. In other words, there exists N such that for all n ≥ N ,
∞
X
nc
ln n < nc . Let bn = 2 . So an < bn for n ≥ N . If
bn converges, then
n
n=1
∞
X
1
an converges by the Direct Comparison Test. Indeed we can choose c =
2
n=1
∞
X
so that
bn converges. All in all, the given series converges.
n=1
(9) (a) (You may use a calculator for this part) If we use Simpson’s rule to approximate
Z 2
ln(x + 1)dx, what is the minimum number of intervals required to ensure
1
that |ES | < 10−4 ?
4
d
Solution: Note that on [1, 2], 4 ln(x + 1) = 6(x + 1)−4 . Since this is a
dx
3
decreasing function on the said interval, it attains maximum 6(1 + 1)−4 =
8
3
at x = 1. So we can take M = .
8
M (b − a)5
< 10−4
180n4
3
5
8 · (2 − 1)
< 10−4
180n4
1
125 4
n>
≈ 2.1364
6
|ES | =
Since n must be an even integer, n is at least 4.
(b) Show that, for any polynomials y = p(x) of degree not exceeding 3, Simpson’s
Z b
Rule gives the exact value of
p(x)dx.
a
13
Solution: Since p(4) (x) is continuous, we have
M (b − a)5
180n4
where M is an upper bound of |p(4) (x)| for a ≤ x ≤ b. But since the degree
of p(x) does not exceeed 3, p(4) (x) ≡ 0 for all x. So we can take M to be 0.
Thus |ES | ≤ 0 =⇒ ES = 0 and Simpson’s Rule always gives the exact value
of the integral.
(c) Let y = f (x) be an odd function. If we use Simpson’s Rule with n = 4
Z 2
f (x)dx, what is |ES |?
subintervals of equal length to approximate
|ES | ≤
−2
Solution: Recall that an odd function satisfies f (−x) = −f (x). Besides its
integral over an interval symmetric about 0 is always 0, i.e.
Z 2
f (x)dx = 0
−2
If we use Simpson’s Rule with n = 4, then x0 = −2, x1 = −1, x2 = 0, x3 = 1,
x4 = 2.
Z 2
2 − (−2)
|ES | = (f (−2) + 4f (−1) + 2f (0) + 4f (1) + f (2)) −
f (x)dx
3(4)
−2
1
= (−f (2) − 4f (1) + 2f (0) + 4f (1) + f (2))
3
2
= f (0)
3
= 0 (f (0) = f (−0) = −f (0) =⇒ f (0) = 0)