Math 113, Calculus II
Final Exam Solutions
1. (25 points) Use the limit definition of the definite integral and the sum formulas
Z
2
−2
1
to compute
4 − x2 dx.
Then check your answer using the Evaluation Theorem.
(2) − (−2)
4
Solution: In this integral, we have a = −2 and b = 2, so ∆x =
=
and
n
n
4
xj = a + j∆x = −2 + j. Thus the limit definition of the definite integral gives
n
Z 2
n
X
2
f (xj )∆x
[4 − x ] dx = lim
n→∞
−2
j=1
n
X
"
# 4 2
4
4 − −2 + j
= lim
n→∞
n
n
j=1
n X
16
4
16 2
4 − 4 − j + 2j
= lim
n→∞
n
n
n
j=1
n X
64 2
64
j
−
j
= lim
n→∞
n2
n3
j=1



 
n
n
X
X
64
64
= lim  2 
j − 3 
j 2 
n→∞ n
n
j=1
j=1
64 n(n + 1)(2n + 1)
64 n(n + 1)
− 3
= lim
n→∞ n2
2
n
6
64
n(n + 1)
n(n + 1)(2n + 1)
64
−
=
lim
lim
2 n→∞
n2
6 n→∞
n3
32
32
= 32(1) − (2) = .
3
3
Now, using the Evaluation Theorem to check our work, we find
Z
2
x3
[4−x ] dx = 4x −
3
−2
2
2
−2
(2)3
(−2)3
8
−8
32
= 4(2) −
− 4(−2) −
= 8 − − −8 −
= .
3
3
3
3
3
2. (25 points) Suppose f is the continuous function given by f (x) =
Rx
0
2
e−t dt.
(a) What is f (0)?
R0
2
Solution: By definition of f , f (0) = 0 e−t dt, so f (0) = 0 because the definite integral
of any function from a to a is always zero.
(b) Is f (1) positive or negative? Support your answer.
R1
2
2
Solution: By definition of f , f (1) = 0 e−t dt, so f (1) is positive because e−t > e−1
on the interval [0, 1], which implies that f (1) ≥ e −1 [1 − 0] = e−1 > 0.
n
1P
i=1
1 = n,
n
P
i=1
i=
n
n(n + 1)(2n + 1)
n(n + 1) P
,
.
i2 =
2
6
i=1
Math 113, Calculus II
Final Exam Solutions
(c) Find f 0 (x).
Solution: By the Fundamental Theorem of Calculus, we have
Z x
d
2
0
−t2
f (x) =
e
dt = e−x .
dx 0
(d) If F (x) = f (sin x), find F 0 (x).
Solution: Using the Chain Rule together with our answer in part (c), we have
F 0 (x) = f 0 (sin x) ·
d
2
2
[sin x] = e−(sin x) · [cos x] = e− sin x cos x
dx
3. (25 points) Consider the function f (x) =
1
on the interval [1, 3].
x
R3 1
1 x dx.
Solution: By the Evaluation Theorem, we have
Z 3
1
dx = [ln |x|]31 = ln |3| − ln |1| = ln 3 − 0 = ln 3
1 x
(a) Compute
R3 1
1 x dx.
3−1
1
1
Solution: With n = 4 subintervals, we have ∆x =
= , and xj = 1 + j for
4
2
2
j = 0, 1, 2, 3, 4. Thus the Trapezoidal Approximation is
Z 3
∆x
1
dx ≈ T4 =
[f (1) + 2f (1.5) + 2f (2) + 2f (2.5) + f (3)]
2
1 x
1/2
67
=
[1 + 2(2/3) + 2(1/2) + 2(2/5) + (1/3)] =
= 1.116666666
2
60
(b) Use the Trapezoidal Rule with n = 4 to approximate
(c) According to the error formula for the Trapezoidal Rule 2 what is the maximum possible
error for this approximation in part (b)? Compare this to the actual error.
Solution: First, with f (x) = x1 = x−1 , we have f 0 (x) = −x−2 and f 00 (x) = 2x−3 . So,
on the interval [1, 3], we have |f 00 (x)| = |2/x3 | ≤ 2, so take M = 2 in the error formula:
|E4 | ≤
16
1
[2](3 − 1)3
=
=
= 0.08333333
2
12(4)
12(16)
12
So the maximum possible error in part (b) is 0.083333, but the actual error is
67
|E4 | = − ln 3 = 0.018054377999
60
which is substantially smaller than the maximum possible error.
2
If |f 00 (x)| ≤ M on [a, b], then the error formula for the Trapezoidal Rule is |En | ≤
M (b − a)3
.
12n2
Math 113, Calculus II
Final Exam Solutions
4. (30 points) The gamma function Γ(x) is defined for all x > 0 by
Z ∞
e−t tx−1 dt.
Γ(x) =
0
(a) Evaluate Γ(1).
Solution: To evaluate Γ(1), we substitute 1 in for x within theR integrand, soRthe expo∞
∞
nent of t is now (1)−1 = 0 and the integral simplifies to Γ(1) = 0 e−t t0 dt = 0 e−t dt.
Thus we have
Z ∞
Z b
h
i
b
e−t dt = lim
e−t dt = lim −e−t 0 = lim −e−b + e0 = 0 + 1 = 1
Γ(1) =
0
b→∞ 0
b→∞
b→∞
(b) For x > 1, show that Γ(x) = (x − 1)Γ(x − 1). [Assume that all these improper integrals
exist and use integration by parts.]
Solution: Following the suggestion, we use integration by parts with
u = tx−1 ,
du = (x − 1)tx−2 dt;
dv = e−t dt, v = −e−t
R
R
Therefore, recalling the formula u dv = uv − v du, we have
Z ∞
Z b
−t x−1
e t
dt = lim
(tx−1 ) (e−t dt)
Γ(x) =
b→∞
0
0
Z b
b
= lim tx−1 (−e−t ) 0 −
(−e−t )[(x − 1)tx−2 ] dt
b→∞
0
Z b
−t
x−2
x−1 −b
x−1 0
e (x − 1)t
dt
= lim
−b e − 0 e +
b→∞
0
Z ∞
e−t tx−2 dt = (x − 1)Γ(x − 1)
= 0 − 0 + (x − 1)
0
(c) Use parts (a) and (b) to find Γ(2), Γ(3), Γ(4). What is the pattern?
Solution: By part (b), we find that
Γ(2) = ([2] − 1)Γ([2] − 1) = 1 · Γ(1) = 1,
Γ(3) = ([3] − 1)Γ([3] − 1) = 2 · Γ(2) = 2 · 1 = 2,
Γ(4) = ([4] − 1)Γ([4] − 1) = 3 · Γ(2) = 3 · (2 · 1) = 6
From just these few computations, we can already see the pattern emerge that
Γ(n) = (n − 1)Γ(n − 1) = (n − 1)[(n − 2)Γ(n − 2)] = · · · = (n − 1)!
5. (20 points) Consider the region R in the first quadrant bounded on the left by the parabola
x = y 2 and on the right by the line y = x.
(a) Sketch the curves and shade the region below:
2
y=x
1
−1
−1
1
2
x = y2
Math 113, Calculus II
Final Exam Solutions
(b) Set up, but do not evaluate!, integrals that compute
i. the area of R;
Solution: Setting it up with respect to x, the parabola is on top and the line is the
bottom, so the area is equal to
Z 1
√
A=
( x) − (x) dx
x=0
If, on the other hand, we integrate with respect to y, then the line is “on top” while
the parabola is the bottom, so the area between them is
Z 1
A=
(y) − (y 2 ) dy
y=0
ii. the volume of the solid obtained by revolving R about the y-axis;
Solution: If we revolve about the y-axis, we should set the integral up with respect
to y. As in part (a), the line x = y is on top and the parabola x = y 2 is on bottom,
so the volume we obtain is given by
Z 1
Z 1
2
2 2
V =
π(y) − π(y ) dt =
π(y 2 − y 4 ) dy
y=0
y=0
iii. the volume of the solid obtained by revolving R about the x-axis.
Solution: If we now revolve the region R about the x-axis, we should write our
√
integral with respect to x, in which case y = x is on top and y = x is on bottom,
so we have
Z 1
Z 1
√ 2
π( x) − π(x)2 dx =
V =
π(x − x2 ) dx
x=0
x=0
6. (25 points) Phoebe Small is caught speeding. The fine is $3.00 per minute for each mile per
hour above the speed limit. Since she was clocked at speeds as much as 64 MPH over a
6-minute period, the judge fines her
($3.00)(number of minutes)(MPH over 55) = ($3.00)(6)(64 − 55) = $162.
Phoebe believes that the fine is too large since she was going 55 MPH at times t = 0 and
t = 6 minutes, and was going 64 MPH only at t = 3. She reckons, in fact, that her speed v
was given by v = 55 + 6t − t2 .
(a) Show that Phoebe’s equation does give the correct speed at times
i. t = 0: v(0) = 55 + 6(0) − (0)2 = 55
ii. t = 3: v(3) = 55 + 6(3) − (3)2 = 64
iii. t = 6: v(6) = 55 + 6(6) − (6)2 = 55
(b) Phoebe argues that since her speed varied, the fine should be determined by her average
speed rather than her maximum speed. What was Phoebe’s average speed?
Math 113, Calculus II
Final Exam Solutions
Solution: Assuming that Phoebe’s speed is given by v(t) = 55 + 6t − t 2 , as we will, her
average speed is
Z 6
Z
1
1 6
vave =
v(t) dt =
55 + 6t − t2 dt
(6) − (0) 0
6 0
6
1
1
t3
(6)3
1
2
2
=
=
55t + 3t −
55(6) + 3(6) −
− (0) = [366] = 61
6
3 0 6
3
6
Thus Phoebe’s average speed is 61 mph, which is still over the speed limit, so she should
still pay a fine.
(c) What should Phoebe propose to the judge as a reasonable fine?
Solution: Based on Phoebe’s average speed, her fine should be
($3.00)(number of minutes)(average MPH over 55) = ($3.00)(6)(61 − 55) = $108.
Thus, if the judge goes along with Phoebe’s argument, her fine would only be $108.
7. (25 points) The Riemann zeta-function ζ is defined by
ζ(x) =
∞
X
1
nx
n=1
and is used in number theory to study the distribution of prime numbers.
P
1
(a) What is the domain of ζ? (That is, for what values of x does ∞
n=1 nx converge?)
Solution: Looking at the right-hand side of the Riemann zeta-function, we recognize
it to be a p-series with p = x. We know that p-series only converge when p > 1
and that they diverge whenever p ≤ 1, so the domain of the Riemann zeta-function is
(1, ∞) = {x : x > 1}.
P
1
(b) Approximate the value of ζ(2) = ∞
n=1 n2 by using the sum of the first 6 terms, s 6 .
Solution: We use the approximation
ζ(2) ≈ s6 =
6
X
1
1
1
1
1
1
5369
1
= 2+ 2+ 2+ 2+ 2+ 2 =
= 1.49138888....
2
n
1
2
3
4
5
6
3600
n=1
R∞
(c) Estimate the error |ζ(2) − s6 | using the error bound Rn ≤ n f (x) dx.
Solution: The error in part (b) is, at worst,
Z ∞
Z b
−1 b
−1 −1
1
1
−2
|ζ(2)−s6 | = R6 ≤
=
dx = lim
x dx = lim −x 6 = lim
−
2
b→∞
b→∞ 6
b→∞
x
b
6
6
6
(d) How large must we take n to insure the error R n = ζ(2) − sn is less than 0.001?
R∞
Solution: To make sure Rn < 0.001, we’ll find n large enough so that n x−2 dx <
0.001. Now
Z ∞
Z b
−1 b
−1 −1
1
−2
−2
x dx = lim
−
x dx = lim −x n = lim
=
b→∞
b→∞ n
b→∞
b
n
n
n
1
< 0.001. Hence we need 1000 < n, so take n = 1001 to ensure that
so we want
n
Rn = R1001 is less than 0.001.
Math 113, Calculus II
Final Exam Solutions
8. (25 points) Consider the series
∞
X
(3x − 2)n
n=1
n
.
(a) Find the series’ radius of convergence.
Solution: To find the radius of convergence, we use the Ratio Test:
(3x − 2)n+1 /(n + 1) an+1 n
= lim lim = lim (3x − 2) = |3x − 2|
n
n→∞
n→∞
n→∞ n + 1
an
(3x − 2) /n
According to the Ratio Test, the power series converges (absolutely) whenever |3x−2| < 1
and diverges when |3x − 2| > 1. Now
|3x − 2| < 1
=⇒
−1 < 3x − 2 < 1
=⇒
1 < 3x < 3
=⇒
1
<x<1
3
has width 2/3, so the radius of convergence is R = 1/3.
(b) Find the interval of convergence of the series.
Solution: So far, we know that the power series converges for 1/3 < x < 1 and diverges
when x < 1/3 or x > 1. It only remains to check the two endpoints:
P
P∞ (3[1/3]−2)n
(−1)n
= ∞
i. x = 1/3: The series at this endpoint is
n=1
n=1
n
n , which is
the Alternating Harmonic Series that we already know to converge (by the Alternating Series Test, in case you don’t remember). So x = 1/3 is in the interval of
convergence.
P
P
(3[1]−2)n
1
= ∞
ii. x = 1: The series at this endpoint is ∞
n=1 n , which is the Harn=1
n
monic Series that we already know to diverge (or you may think of it as p-Series
with p = 1 ≤ 1). Thus x = 1 is not in the interval of convergence.
Therefore, the interval of convergence is [1/3, 1) = {x : 1/3 ≤ x < 1}.
(c) Where does the series converge absolutely? Where does it converge conditionally?
Solution: According to the Ratio Test, we know that the series converges absolutely
on the interval (1/3, 1) = {x : 1/3 < x < 1}. By part (b), we also know that the series
converges at x = 1/3 but not at x = 1, so it is only conditionally convergent at x = 1/3.
Thus the power series
• converges absolutely on (1/3, 1)
• converges conditionally at x = 1/3
• diverges whenever x < 1/3 or x ≥ 1
9. (25 points) Use substitution, multiplication, differentiation, and/or integration to find Taylor
series centered at a = 0 for the following functions:
(a)
x
1 + x2
Solution:
(b) cos(2x)
P
P∞
P∞
x
1
2 n
n 2n =
n 2n+1
= x·
= x· ∞
n=0 (−x ) = x· n=0 (−1) x
n=0 (−1) x
2
2
1+x
1 − (−x )
Math 113, Calculus II
Final Exam Solutions
x2n+1
d P∞
x2n+1
d
n
, cos x = dx [sin x] =
=
Solution: Since sin x = n=0
n=0 (−1)
(2n + 1)!
dx
(2n + 1)!
2n
2n
2n
P∞
P∞
P∞
n (2n + 1)x
n x
n (2x)
(−1)
=
(−1)
.
Thus
cos(2x)
=
(−1)
=
n=0
n=0
n=0
(2n + 1)!
(2n)!
(2n)!
2n 2n
2n
P∞
P∞
n2 x
n x
(−1)
=
(−4)
.
n=0
n=0
(2n)!
(2n)!
P∞
(−1)n
(c) tan−1 x
Solution: We have
P∞
n=0 (−1)
(d)
tan−1 x
x
Solution:
(e) xex
2
n
x2n+1
2n + 1
tan−1
x=
.
Rx
0
2n+1 x
R x P∞
P∞
t
1
2
n
n
=
dt = n=0 (−1)
dt = 0
n=0 (−t )
1 + t2
2n + 1 0
2n+1
2n
P
P
tan−1 x
n x
n x
= x−1 · tan−1 x = x−1 · ∞
= ∞
.
n=0 (−1)
n=0 (−1)
x
2n + 1
2n + 1
2
Solution: xex = x ·
P∞
n=0
P
(x2 )n
x2n P∞ x2n+1
=x· ∞
= n=0
.
n=0
n!
n!
n!
10. (25 points) Find the Taylor polynomial of order 4 (i.e., only up to the (x − a) 4 -th term)
generated by f (x) = sin x centered at a = π/3.
Solution: In order to find the Taylor polynomial of order 4 centered at a = π/3 for f (x) =
sin x, which is
4
X
f (n) (a)
(x − a)n
T4 (x) =
n!
n=0
we need to know the first 4 derivatives of f (x) and their values at a = π/3. So we find
√
f (x) = sin x
=⇒
f (π/3) = sin(π/3) = 3/2
f 0 (x) = cos x
=⇒
f 00 (x) = − sin x
=⇒
f 000 (x) = − cos x
f (4) (x) = sin x
=⇒
=⇒
f 0 (π/3) = cos(π/3) = 1/2
√
f 00 (π/3) = − sin(π/3) = − 3/2
f 000 (π/3) = − cos(π/3) = −1/2
√
f (4) (π/3) = sin(π/3) = 3/2
Thus, the Taylor polynomial of order 4 generated by f (x) = sin x centered at a = π/3 is
T4 (x) =
4
X
f (n) (π/3) n=0
n!
x−
π n
3
f 0 (π/3) π 1 f 00 (π/3) π 2 f 000 (π/3) π 3 f (4) (π/3) π 4
= f (π/3) +
x−
x−
x−
x−
+
+
+
1!
3√
2!
3
3!
4!
3
√ 3
√
3 1/2 3/2
π − 3/2 π 2 −1/2 π 3
π 4
+
x−
+
x−
+
x−
+
x−
=
1
3 √ 2
3
6
3
24
3
√
√2
2
3
4
3 1
3
3
π
π
1
π
π
+
x−
−
x−
−
x−
+
x−
=
2
2
3
4
3
12
3
48
3