Calculus II Final Exam Problems - Solutions 1 1. Use the definition of ...

Calculus II Final Exam Problems - Solutions 1
1. Use the definition of the definite integral to evaluate 1
∫ 1
−1
(3x − x 2 ) dx.
Solution: Recall the limit definition of the definite integral:
∫ b
a
f(x) dx = lim
n→∞
i=1
n∑
f(x ∗ i )∆x
where ∆x = b−a
n , x i = a + i∆x and x ∗ i is an arbitrary point between x i−1 and x i . In
our case above, f(x) = 3x − x 2 , a = −1, and b = 1. Thus, for a particular value of n,
∆x = (1)−(−1)
n
= 2 n and x i = −1 + i 2 n . Therefore, using x∗ i = x i for i = 1, 2, . . . , n,
∫ 1
−1
(3x − x 2 ) dx = lim
n→∞
i=1
= lim
n→∞
= lim
n→∞
= lim
n→∞
2
= lim
n→∞ n
2
= lim
n→∞ n
n∑
f(x i )∆x
n∑
[3(x i ) − (x i ) 2 ] 2 n
i=1
n∑
[
3(−1 + 2i
]
2i 2
) − (−1 +
n n )2 n
i=1
n∑
[
(−3 + 6i
]
) − (1 − 22i
n n + 4i2 2
n 2 ) n
i=1
n∑
[−4 + 10 n i − 4 ]
n 2 i2
i=1
[
−4(
]
n∑
1) + 10 n n ( ∑
i) − 4 n n 2 ( ∑
i 2 )
i=1
i=1
[
2
= lim −4(n) + 10 ( n(n + 1)
n→∞ n
n 2
[
= lim (−8) + 20 n(n + 1)
n→∞ n 2 2
= lim
n→∞
i=1
)
− 4 ( n(n + 1)(2n + 1)
n 2 6
]
− 8 n 3 n(n + 1)(2n + 1)
6
[
n(n + 1)
(−8) + 10
n 2 − 4 ]
n(n + 1)(2n + 1)
3 n 3
= (−8) + 10(1) − 4 3 (2) = 2 − 8 3 = −2 3 .
To check that this answer is correct, we can use the Fundamental Theorem of Calculus:
∫ 1
−1 (3x − x2 ) dx = [3 x2
2 − x3
3 ]1 −1 = [(3 1 2 − 1 3 ) − (3 1 2 − −1
3 )] = − 2 3 .
2. The Sierpinski Triangle is constructed in the following way: Start with an equilateral triangle
of side length s and area A. Remove the equilateral triangle which has its vertices
at the midpoint of each of the original sides from the center of the original triangle. What
)]
1
n∑
i =
i=1
n(n + 1)
2
n∑
i 2 =
i=1
n(n + 1)(2n + 1)
6
n∑
i=1
i 3 = n2 (n + 1) 2
4

Calculus II Final Exam Problems - Solutions 2
remains is a collection of three triangles, each with side length s/2 and area A/4. Perform
the center-removing operation on each of these three triangles. The result will be 9 triangles
of side length s/4. Remove the centers of these 9 triangles, and so on. . . . See the figure.
(a) Use geometric series arguments to show that the total area removed is A, i.e. the area
of the Sierpinski triangle is 0.
Solution: Let A be the area of the original equilateral triangle. In the first step
toward constructing the Sierpinski Triangle, we remove 1 triangle whose area is A/4,
leaving three triangles each with area A/4. From each of these three, we remove the
middle triangle of area 1/4 the size of the triangle from which it was removed and so
produce three new triangles left with area A/16 = A/4 2 . Notice that the 3s counting
the number of triangles that remain compound as we go from stage to stage, for in
each step we remove the middle triangle from all of the ones we currently have and
thus produce 3 from the one for each triangle. So we start with 1 triangle and get 3,
from the 3 we get 9, from the 9 we’ll get 27, and in general there will be 3 n triangles
at the end of the nth step and we remove one from each of these in the n + 1st step, so
we remove 3 n−1 triangles in step n. The areas of these triangles decreases, however,
starting with A/4 for the area of the first triangle we remove, then one fourth of that
for each of the triangles removed in the next step (there are three of them, so we
remove an area equal to 3A/16 in the second step), then one fourth of the area per
triangle (so now A/64 = A/4 3 ) in the third step and there are 9 such triangles, so we
removed the area 9A/64 = 3 2 A/4 3 in the third step) and so forth. Therefore, at the
nth step, we remove 3 n−1 triangles, each with area A/4 n , so the total area removed
is
∞∑
3 n−1 A ∞ 4 n = ∑
( ) n−1
A 3
= A/4
4 4 1 − (3/4) = A/4
1/4 = A.
n=1
n=1
(b) Show that the length of the boundary of the Sierpinski triangle is infinite.
Solution: As noted above, at the nth step, there are 3 n−1 triangles and the perimeters
only decrease by half each time: 3s to start with, then 3s/2 for each triangle of the 3 in
the second figure, (3s/2)/2 = 3s/4 per triangle in the third, and 3s/2 n−1 per triangle
in the nth. So at the nth stage, there are 3 n−1 triangles, each with perimeter equal
to 3s/2 n−1 , so the perimeter is the product, 3 n−1 (3s/2 n−1 ) = 3s( 3 2 )n−1 . Hence the
total length of the Sierpinski Triangle is limit of these lengths: lim n→∞ 3s ( 3 n−1,
2)
which is infinite since 3/2 > 1 so that (3/2) n−1 → ∞ as n → ∞. Therefore, the total
length of the boundary of the Sierpinski Triangle is infinite.
3. Use integration by substitution to show that ∫ f( √ x) dx = ∫ 2uf(u) du. Calculate
∫
sin
√ x dx.
Solution: Let u = √ x = x 1/2 , so du = 1 2 x−1/2 dx. Thus dx = (2x 1/2 ) du = 2 √ x du =

Calculus II Final Exam Problems - Solutions 3
2u du expresses dx in terms of just us and du. Therefore, by substitution,
∫
f( √ ∫
∫
x) dx = f(u) [2u du] = 2uf(u) du.
Applying this, we find ∫ sin √ x dx = ∫ 2u sin u du. To compute the latter, use integration by
parts with w = 2u (so dw = 2 du) and dv = sin u du (so v = − cos u): ∫ 2u sin u du = wv −
∫
v dw = [2u][− cos u]−
∫
[− cos u][2 du] = −2u cos u+2
∫
cos u du = −2u cos u+2 sin u+C.
Therefore,
∫
sin √ x dx = −2u cos u + 2 sin u + C = −2 √ x cos √ x + 2 sin √ x + C.
Not that we need to, as we’re all experts at integration by now, but remember that we can
always check our work by differentiating. So consider d
dx [−2√ x cos √ x + 2 sin √ x]:
= −2[( √ x)(− sin √ x · [1/2x −1/2 ]) + (cos √ x)(1/2x −1/2 )] + 2 cos √ x · [1/2x −1/2 ]
= −2[− 1 2 sin √ x + 1 2 x−1/2 cos √ x] + x −1/2 cos √ x = sin √ x.
This is the integrand of the original function, and that’s what we must get if we did things
right, so we did things correctly.
4. A mathematical operation that takes a function as input and produces another function
as output is sometimes referred to as a transform. An important example is the Laplace
transform L, defined by
L[f(x)] =
∫ ∞
0
e −px f(x) dx = F (p).
Notice that L turns a function of x into a function of the parameter p. Find the Laplace
transform of the functions
(a) f(x) = 1,
(b) g(x) = x and
(c) h(x) = e 5x
(assume in the first two cases that p > 0 and in the third case p > 5).
Solution: The Laplace transform of f(x) = 1 with p > 0 is
L[1] =
∫ ∞
0
e −px [1] dx = lim
b→∞
∫ b
The Laplace transform of g(x) = x with p > 0 is
L[x] =
∫ ∞
0
0
e −px dx = lim [ 1
b→∞ −p e−px ] b 1
0 = lim
b→∞ −p [e−pb − e 0 ] = 1 p .
e −px [x] dx = lim b→∞ [(x)( 1
−p e−px ) − ∫ 1
−p e−px dx] b 0 = lim b→∞[ 1
−p xe−px 1
p 2 e −px ] b 0
= lim b→∞ [( 1
−p be−bp − 1 p 2 e −pb ) − ( 1
−p 0e0 − 1
p 2 e 0 )] = 1
p 2
using integration by parts with u = x (so du = dx) and dv = e −px (so v = 1
−p e−px ) and
l’Hôpital’s Rule to find that lim x→∞ xe −px = 0. Lastly, the Laplace transform of h(x) = e 5x
for p > 5 is
L[e 5x ] =
∫ ∞
0
e −px [e 5x ] dx = lim b→∞
∫ b
0 e(5−p)x dx = lim b→∞ [
1
5−p e(5−p)x ] b 0
= lim b→∞
1
5−p [(e(5−p)b ) − (e 0 )] = −1
5−p = 1
p−5 .

Calculus II Final Exam Problems - Solutions 4
5. Find the limit:
lim
h→0
∫ 1+h √
x5 + 8 dx
1
.
h
√
x5 + 8 dx = F (1 + h) − F (1) and
√
t5 + 8 dt. Then ∫ 1+h
1
Solution: Let F (x) = ∫ x
0
F ′ (x) = √ x 5 + 8 by the Fundamental Theorem of Calculus, so
lim
h→0
∫ 1+h √
x5 + 8 dx
1
F (1 + h) − F (1)
= lim
= F ′ (1) = √ (1)
h
h→0 h
5 + 8 = 3.
6. Starting with
find the sum of the series:
∞∑
f(x) = x n ,
n=0
(a)
(b)
(c)
(d)
(e)
∞∑
nx n−1 , |x| < 1
n=1
∞∑
n=1
n
2 n
∞∑
n(n − 1)x n , |x| < 1
n=2
∞∑
n=2
n 2 − n
2 n
∞∑
n=1
Solution: This is Problem 36 from Section 8.6.
The first sum is related to f(x) by differentiating (nx n−1 = d/dx[x n ]), so
n 2
2 n
∞∑
nx n−1 = f ′ (x), |x| < 1.
n=1
The second sum is related to the first by multiplying by x and then evaluating when x = 1/2,
so
∞∑ n
2 n = 1 ∞∑
[n(1/2) n−1 ] = 1 2
2 f ′ (1/2).
n=1
n=1
The third is just x 2 times the derivative of the first, so
∞∑
n(n − 1)x n = x 2
n=2
∞ ∑
n=2
d
dx [nxn−1 ] = x 2 f ′′ (x), |x| < 1.

Calculus II Final Exam Problems - Solutions 5
The 4th is the value of the third when x = 1/2, so it is
∞∑
n=2
n 2 − n
2 n = [1/2] 2 f ′′ (1/2) = [1/4]f ′′ (1/2).
The last is the sum of the second and the 4th, since in the 4th when n = 1 the term is 0,
so ∑ ∞
n=2 = ∑ ∞
n=1
for that expression, which implies
∞∑
n=1
n 2 ∞
2 n = ∑
n=1
n 2 − n
2 n + n 2 n = ∞ ∑
n=1
n 2 − n
2 n +
∞∑
n=1
n
2 n = [1/4]f ′′ (1/2) + [1/2]f ′ (1/2).
7. Use either a direct comparison or the limit comparison test to decide the convergence or
divergence of each of the following:
∞∑
n=1
1
√
1 + n
5
∞∑
n=3
n ∑
∞ 1
(n + 3) 2 2n 2 − n
n=1
∞∑
n=1
5 sin 2 n
n √ n
Solution: The first series, ∑ √ 1
1+n
, converges by the Direct Comparison Test since
5
1
√
1 + n
5 ≤ 1 √
n
5 = 1
n 5/2
and the series ∑ 1
is a p-series with p = 5/2 > 1 so that it converges.
n 5/2
The second series, ∑ n
(n+3)
, diverges by the Limit Comparison Test since
2
lim
n→∞
n
(n+3) 2
1
n
n 2
= lim
n→∞ (n + 3) 2 = 1 > 0
and the series ∑ 1
n
diverges (p = 1 ≤ 1).
The third series, ∑ 1
2n 2 −n
, converges by the Limit Comparison Test since
lim
n→∞
1
2n 2 −n
1
n 2
n 2
= lim
n→∞ 2n 2 − n = 1 2 > 0
and ∑ 1
n 2 converges (p = 2 > 1).
The last series converges by the Direct Comparison Test because 0 ≤ sin 2 x ≤ 1 implies
that 0 ≤ 5 sin 2 n ≤ 5 for all n, so that
5 sin 2 n
n √ n ≤ 5
n √ n = 5
n 3/2
and the series ∑ 5
n 3/2 = 5 ∑ 1
n 3/2 converges (p = 3/2 > 1).
8. Consider a thin, homogeneous, rigid rod of length L cm and density d g/cm, aligned on the
x-axis with one end fixed at the origin.
Now suppose the rod is rotating around the y-axis with angular velocity ω rev/sec, like a
helicopter rotor. Set up and calculate the integral to find the total kinetic energy of the
rotating rod.

Calculus II Final Exam Problems - Solutions 6
The procedure: Go back to basic principles. Dissect the rod into small pieces, figure out
the kinetic energy of each small bit, and then add all the contributions to the total kinetic
energy with an appropriate integral. The kinetic energy of a point mass m is 1 2 mv2 where
v is the linear velocity, measured in units of cm/sec. The linear velocity of a point on the
rotating rod r units away from the origin is 2πrω.
Solution: Suppose we divide the rod into n parts of equal length, and thus equal mass
since the rod is homogeneous of density d g/cm. Then each part has length ∆x = L n and
mass d · ∆x = d L n = dL n . Now select points x∗ i in each part of the rod, x i−1 ≤ x ∗ i ≤ x i
where x i = 0 + i∆x = iL n . In fact, take x∗ i = x i. Then the kinetic energy of the rotating
rod is approximated by the sum of the kinetic energies of the rotating points x i , with the
approximation improving as we let n → ∞, so the total kinetic energy will be
n∑
KE = lim KE(x i )
n→∞
= lim
n→∞
i=1
n∑ 1
2 (d∆x) v(x i) 2
i=1
d
= lim
n→∞ 2
d
= lim
n→∞ 2
n∑
(2πx i ω) 2 ∆x
i=1
n∑
( ) 2 iL
[4π 2 ω 2 ][ L n n ]
i=1
= lim
n→∞ 4π2 ω 2 dL3
2n 3
∑
n i 2
i=1
[ ]
= 2π 2 ω 2 dL 3 1 n(n + 1)(2n + 1)
lim
n→∞ n 3 6
[ ] 2
= 2π 2 ω 2 dL 3 = 2 6 3 π2 ω 2 dL 3 .
Observe that all of this work was really computing the definite integral ∫ L
d
2 [4π2 ω 2 ] ∫ L
0 x2 dx = 2π 2 ω 2 d[ x3
3
9. Evaluate the definite integral
L
0 = 2π2 ω 2 d[ L3
3 ].
∫ 1
0
f(x) dx
where f is the function whose graph is shown below
0
1
2 (d)(2πxω)2 dx =
1
...
1/8 1/2
1/16 1/4 1
Solution: Remembering that the definite integral of a non-negative function measures the
area under the curve, we can compute this integral geometrically since we know the area

Calculus II Final Exam Problems - Solutions 7
formula for a triangle: A = 1 2bh. In all of the triangles in the graph of y = f(x), the height
h = 1, but the base decreases: b = 1/2 to b = 1/4 to b = 1/8 and so on from the rightmost
triangle working to the left. So
∫ 1
0
f(x) dx =
∞∑
n=1
1
2 [(1/2)n ][1] =
∞∑
( ) n+1 1
=
2
n=1
1/4
1 − (1/2) = 1 2
because the inital term (n = 1) is the area of the first triangle, which is a = 1/2[1/2][1] = 1/4
and the ratio is r = 1/2 (what happens to the width of the triangles) and this is a geometric
series.
10. If F (x) = ∫ 3
0 t√ t + 9 dt, then F ′ (1) = 0. Why?
Solution: The definite integral ∫ 3
0 t√ t + 9 dt measures the area under the curve y =
x √ x + 9 over the interval [0, 3], and this is some constant. Thus F (x) is a constant function,
and the derivative of a constant is 0, so F ′ (x) = 0 for every x. In particular, F ′ (1) = 0.
11. Suppose you have a large supply of books, all the same size, and you stack them at the
edge of a table, with each book extending farther beyond the edge of the table than the
book beneath it: the top book extends half its length beyond the second book, the second
book extends a quarter of its length beyond the third, the third extends one sixth of its
length beyond the fourth, and so on.
(a) Show, by considering the center of mass of the stack, that it is possible to stack the
books so that the top book extends entirely beyond the table. Work from the top
down. First find the center of mass of two books. Then use this result to find the
center of mass when a third book has been added to the bottom of the stack.
Solution: Don’t worry about this part!! We didn’t cover center of mass.
(b) How many books does it take before the top book extends completely beyond the edge
of the table?
Solution: Call the length of the books L, as they are all the same size. Assuming
the result of part (a), that we can make this stack, we want to find the number N so
that the total overhang is L, the length of one book. The amount of overhang from
the top book is L/2, the second is L/4, the third is L/6, and, in general, the nth is
L/(2n). So we want to find N so that ∑ N
n=1 L 2n ≥ L but ∑ N−1
n=1 L 2n
< L. Factoring
out the L on the left hand side and cancelling it on each side, we want to find the first
N with ∑ N
n=1 1
2n = 1 ∑ N
2 n=1 1 n ≥ 1, so ∑ N
n=1 1 n
≥ 2. Starting at N = 1, the sum is
s 1 = 1 1 = 1, which is too small. The partial sum s 2 = 1 1 + 1 2 = 3 2
is also too small, as
is s 3 = 1 1 + 1 2 + 1 3 = 11
6 , but s 4 = 1 1 + 1 2 + 1 3 + 1 4 = 25
12
> 2. Therefore, N = 4 works, so
it only takes 4 books before the top book extends completely beyond the edge of the
table.
(c) By considering the harmonic series, show that the top book can extend any distance
at all beyond the edge of the table if the stack is high enough.
Solution: As we can see in our solution to part (b), the distance that the top book
extends beyond the edge of the table is
N∑
n=1
L
2n = L 2
Since the harmonic series ∑ ∞
n=1 1 n
diverges, its sequence of partial sums is not bounded,
so for any distance D there must be an integer N so that L ∑ N
2 n=1 1 n
> D. (If not, then
N∑
n=1
1
n .

Calculus II Final Exam Problems - Solutions 8
the sequence of partial sums of the harmonic series would be increasing and bounded,
so it would converge.)
12. Here is a “proof” that 1=0. Your job is to find the bug.
On the one hand,
2
π
∫ π
0
∫ π
cos 2 θ dθ = 2 1 + cos 2θ
dθ
π 0 2
= 2 ( 1
π 2 θ + 1 )
4 sin 2θ | π 0
= 2 π
π
2 = 1.
On the other hand, let u = sin θ. Then cos θ = √ 1 − sin 2 θ = √ 1 − u 2 and du = cos θ dθ,
so
2
π
∫ π
0
cos 2 θ dθ = 2 π
= 2 π
∫ π
0
∫ sin π
sin 0
(cos θ)(cos θ) dθ
√
1 − u2 du = 0.
∫
2 π
Solution: The first computation is correct,
π 0 cos2 θ dθ = 1. The second makes the
mistake by assuming cos θ = √ 1 − sin 2 θ as if cos θ were always positive on the interval
[0, π], which it isn’t. Technically, √ cos 2 θ = | cos θ|, which is only equal to cos θ when
cos θ ≥ 0. However, on the interval [0, π], cos θ ≥ 0 only for 0 ≤ θ ≤
√ π 2 . When π 2 ≤ θ ≤ π,
1 − sin 2 θ = | cos θ| = − cos θ.
13. (a) Use Taylor’s Theorem to find the Maclaurin series for f(x) = e x .
Solution: The Maclaurin series of a function f(x) is its Taylor series centered at
x = 0, so we know this to be
∞∑
n=0
x n
n! = 1 + x + x2
2! + x3
3! + · · · + xn
n! + · · · .
The way this is derived using Taylor’s Theorem, which is the formula
T (x) =
∞∑
n=0
f (n) (a)
(x − a) n ,
n!
is by realizing that f (n) (x) = e x for all n ≥ 0 since e x is its own derivative, and that
e 0 = 1 implies that every coefficient will be 1 n! .
(b) Use part (a) to find the Maclaurin series for e −x3 . What is the interval of convergence
of this series? For what x does it converge absolutely?
Solution: By substitution, we have
∞∑ (−x 3 ) n ∞∑ (−1) n x 3n ∞∑
e −x3 =
=
=
n!
n!
n=0
n=0
n=0
(−1) n x3n
The radius of convergence for e x = ∑ x n
n!
is R = ∞, so that the series converges
absolutely for all |x| < ∞. Thus the series we obtained for e −x3 by substitution will
n! .

Calculus II Final Exam Problems - Solutions 9
converge absolutely whenever | − x 3 | = |x 3 | < ∞, which again is for all |x| < ∞.
Hence the interval of convergence is R = (−∞, ∞), on which the series converges
absolutely.
(c) Calculate
∫ 0.2
0
e −x3 dx
to 4-decimal place accuracy. Explain how you know that your answer is sufficiently
accurate.
Solution: Don’t worry about this one, as it requires the error bound on Taylor’s
Theorem, that the remainder is of the form
R n (x) = f (n+1) (z)
(x − a)n+1
(n + 1)!
for some z between a and x. All you need to do is find an n sufficiently large so that
for every value of z and of x between 0 and 0.2, the value of R n (x) ≤ M for some
constant M such that ∫ 0.2
0
M dx = M(0.2 − 0) = 0.2M < 0.00005 (so M < 0.00025).
The reason this works is that, for this value of n, Taylor’s Theorem states that
f(x) = f(a)+ f ′ (a)
1!
(x−a)+ f ′′ (a)
2!
(x−a) 2 +· · ·+ f (n) (a)
(x−a) n +R n (x) = T n (x)+R n (x).
n!
Hence ∫ 0.2
0
f(x) dx = ∫ 0.2
0
[T n (x) + R n (x)] dx = ∫ 0.2
0
T n (x) dx + ∫ 0.2
0
R n (x) dx. Therefore
| ∫ 0.2
f(x) dx − ∫ 0.2
T
0 0 n (x) dx| = | ∫ 0.2
R
0 n (x) dx| ≤ ∫ 0.2
M dx = 0.2M < 0.00005.
0
This means that ∫ 0.2
T
0 n (x) dx approximates ∫ 0.2
f(x) dx accurately to within 4 decimal
0
places.