Solutions to Practice Problems

Math 113, Calculus II
Test 1: Review Solutions
1. Use the limit definition of the definite integral to evaluate
∫ 5
2
(4x − x 2 )dx.
Solution: By definition, ∫ 5
2 (4x − ∑ x2 )dx = lim n
n→∞ i=1 f(x∗ b−a
i )∆x, where ∆x =
n
= 5−2
n =
3
n , f(x) = 4x − x2 and we may take x ∗ i = x i to be the right-hand endpoints of the n
subintervals, which are given as x i = a + i∆x = 2 + i 3 n = 2 + 3i
n . Thus,
∫ 5
2
(4x − x 2 )dx = lim
n→∞
= lim
n→∞
= lim
n→∞
= lim
n→∞
= lim
n→∞
n∑
i=1
n∑
i=1
n∑
i=1
n∑
i=1
n∑
i=1
= lim
n→∞ [(12 n
= lim
n→∞ [12
f(2 + 3i
n ) · 3
n
[4(2 + 3i
n
[8 + 12i
n
[4 − 9i2
n 2 ] · 3
n
[ 12
n − 27i2
n 3 ]
n∑
i=1
= lim
n→∞ [12 − 27 6
) − (2 +
3i
n )2 ] · 3
n
− (4 +
12i
n + 9i2
n 2 )] 3 n
1) − ( 27
n 3
n∑
i 2 )]
i=1
27 + 1)(2n + 1)
(n) − (n(n )]
n n3 6
(n + 1)(2n + 1)
n 2 ]
= 12 − 27
6 lim (n + 1
n→∞ n )(2n + 1 )
n
= 12 − 9 [(1)(2)] = 12 − 9 = 3.
2
Hence, ∫ 5
2 (4x−x2 )dx = 3. [Here’s a quick check using FTC(II): ∫ 5
2 (4x−x2 )dx = 4 x2
(50 − 125
3 ) − (8 − 8 117
3
) = 42 −
3
= 42 − 39 = 3.]
2 − x3
3 |5 2 =
2. There is a rule of thumb that the area of any parabolic arch is two-thirds the product of
its height h and the length of its base b. Confirm this formula by using the Fundamental
Theorem of Calculus on a suitable function to calculate the area in question.
Solution: The function f(x) = 4h (bx − x 2 ) has graph which is a parabola passing through
b 2
the points (0, 0), (b, 0) (so that its base is b) and height h = f(b/2). Thus we want to show
that ∫ b
0 f(x)dx = 2 3bh. So consider
∫ b
0
as desired.
4h
b 2 (bx − x2 )dx = 4h ∫ b
b 2 (bx − x 2 )dx = 4h
b 2 [bx2 2 − x3
3 ]b 0 = 4h
b 2 [(bb2 2 − b3 3 ) − (0)]
0
= 4h
b 2 [b3 ( 1 2 − 1 3 )] = 4hb[1 6 ] = 2 3 bh

Math 113, Calculus II
Test 1: Review Solutions
3. Suppose you know that a certain function f is twice differentiable and that its graph over
[0, 2] is given in the figure on your review sheet. As you see, the printer was sloppy and spilled
a lot of ink on the graph. Decide, if possible, whether each of the following definite integrals
is positive, equal to zero, or negative.
(a) ∫ 2
0 f ′′ (x)dx
(b) ∫ 2
0 f ′ (x)dx
(c) ∫ 2
0 f(x)dx
Solution: By the Fundamental Theorem of Calculus (Part I), we know that ∫ 2
0 f ′′ (x)dx =
f ′ (2) − f ′ (0) because f ′ (x) is an antiderivative of f ′′ (x). Now f ′ measures the slope of the
curve, and we see that the slope of f is negative when x = 2 (because the curve is decreasing
there!) so that f ′ (2) < 0. Similarly, since f is increasing when x = 0, f ′ (0) > 0, so
f ′ (2) − f ′ (0), whatever the actual value of the numbers may be, must be negative as it is a
negative number (f ′ (2)) MINUS a positive number (f ′ (0)). Thus,
∫ 2
0
f ′′ (x)dx < 0.
Using the Fundamental Theorem of Calculus (Part I) again, ∫ 2
0 f ′ (x) = f(2)−f(0) since
f(x) is an antiderivative of its derivative, f ′ (x). Looking at the graph of y = f(x), we see
that f(2) is a small positive number while f(0) is a large positive number, so that f(2) − f(0)
must then be negative. Therefore,
∫ 2
0
f ′ (x)dx < 0.
Lastly, we can see a fair bit of the graph, and the portion of it to the left of the ink spot is all
well above the x-axis and by itself contributes a positive area larger than the total area of the
ink blot BELOW the x-axis. Thus, as ∫ 2
0
f(x)dx measures the signed area of f(x) over the
interval [0, 2] and the part of this area to the left of the ink blot is larger than any negative
amount the ink blot could be covering and the part to the right of the blot is again positive,
we conclude that the signed area must be positive, so
4. If f is a continuous function, find f(4) if
∫ 2
0
x sin πx =
f(x)dx > 0.
∫ x 2
0
f(t)dt.
Solution: Let F (x) = ∫ x
0
f(t)dt, as in the statement of the Fundamental Theorem of
Calculus (Part I). Then, applying the theorem, F ′ (x) = f(x). Now, the integral on the
right-hand side of the equation above is not F (x), but rather is F (x 2 ), for the upper limit of
integration is x 2 instead of simply x. Therefore, the derivative of the right-hand side is
d
dx [F (x2 )] = F ′ (x 2 d
) ·
dx [x2 ] = f(x 2 ) · [2x] = 2x · f(x 2 )

Math 113, Calculus II
Test 1: Review Solutions
because F ′ = f and using the chain rule. The derivative of the left-hand side, which must
equal the function above, is found using the product rule to be
Thus 2xf(x 2 ) = πx cos(πx) + sin(πx), so that
(x)[cos(πx) · π] + (sin πx)[1].
f(x 2 ) = 1 [πx cos(πx) + sin(πx)].
2x
To find the value of f(4), we need to know the value of x so that x 2 (the input of our only
known formula for f above) is equal to 4, i.e., sove x 2 = 4. Hence we may take x = 2 (we
could take x = −2 and would get the same answer - check this!) to find
f(4) = f(2 2 ) = 1
2(2) [π(2) cos(π · 2) + sin(π · 2)] = 1 4 [2π(1) + (0)] = π 2 .
5. Steve and Rick are having and argument as to the value of
∫
sec 2 x tan xdx.
Steve makes the substitution u = sec x and Rick uses the substitution v = tan x. Both
substitutions seem to work, but they arrive at completely different-looking answers. What’s
going on?
Solution: Let’s start with Steve’s substitution, u = sec x, in which case du = sec x tan xdx
and
∫
∫
∫
sec 2 x tan xdx = sec x · sec x tan xdx = udu = u2
2 + C = 1 2 sec2 x + C
for some arbitrary constant C. In contrast, using Rick’s substitution, v = tan x, we have
dv = sec 2 xdx, so that
∫
∫
sec 2 x tan xdx = vdv = v2
2 + D = 1 2 tan2 x + D
for some arbitrary constant D. Now we can tell why Steve and Rick are upset with one
another! These two things don’t look alike, yet they are!!! Recall the trigonometric identity
involving sec 2 x and tan 2 x that states tan 2 x + 1 = sec 2 x. Therefore, if we replace the sec 2 x
in Steve’s solution using this identity, we have
1
2 sec2 x + C = 1 2 (tan2 x + 1) + C = 1 2 tan2 x + 1 2 + C = 1 2 tan2 x + D
if we take Rick’s D to be Steve’s C + 1 2
, which we can do since the constants are all arbitrary.
6. Suppose you and two friends order a 14-inch diameter pizza and decide to split it up among
you by sectioning it as shown in the figure on your review sheet. If you are really hungry,
which slice do you ask for? Justify your answer.
Solution: Let’s begin by placing the origin at the center of the circle, so that we can exploit
as much symmetry as possible as well as write the equation for the circle as x 2 +y 2 = 7 2 = 49.
If we are particularly hungry (I’m not right now, but this might be good to know later), we’ll
want to select the one of the three pieces with the largest area. Now, of course, when we

Math 113, Calculus II
Test 1: Review Solutions
4
∫ 2
0
hear the term “area”, we must think of integrals, so we want to find the largest area using
integration.
Looking at the picture of the pizza, if we can figure out the area under the circle, above the
x-axis, and between the lines x = 0 and x = 2, this is one-fourth of the total area of the
middle piece. So the area of the middle piece is
√
49 − x 2 dx = 4[ x √
49 − x
2 2 + 49
2 sin−1 ( x 7 )]2 0 = 4[( 2 √ 49 2 49 − 4+
2 2 sin−1 )−(0)] = 55.22848247....
7
As the total area of the pizza is π(7) 2 = 153.93804 and the two end pieces are of the same
area, each must have area 1 2
(153.93804 − 55.22848247) = 49.35477878, which is less than the
area of the middle piece. Hence, if we are both hungry and selfish (hunger alone shouldn’t be
an excuse for taking the largest piece), we should ask for the middle slice.
7. The average y-value of a function y = f(x) over the interval [a, b] is equal to the number
1
b − a
∫ b
a
f(x)dx.
[We will see later where this formula comes from and why it gives the average y-value.]
(a) Find the average values of f(x) = x, f(x) = x 2 , f(x) = x 3 , and f(x) = x 4 on the interval
[0, 1].
Solution: For all of these functions, b = 1, a = 0, so b − a = 1 − 0 = 1 and the average
value is therefore simply 1 ∫ 1
1−0 0 f(x)dx = ∫ 1
0
f(x)dx. Thus, the average value of f(x) = x
is ∫ 1
x2
0
xdx = [
2 ]1 0 = 1/2; the average value of f(x) = x2 is ∫ 1
0 x2 dx = [ x3
3 ]1 0 = 1/3; the
average value of f(x) = x 3 is ∫ 1
0 x3 dx = [ x4
4 ]1 0 = 1/4; and the average value of f(x) = x4
is ∫ 1
0 x4 dx = [ x5
5 ]1 0 = 1/5.
(b) From the pattern that is established in the previous part, what would you guess is the
average value of f(x) = x n on the interval [0, 1] for any integer n ≥ 1? Justify your
guess.
Solution: We appear to have the average value of f(x) = x n on the interval [0, 1] equal
to 1/(n + 1). Let’s see if that is always correct. The average value of f(x) = x n is
∫ 1
0 xn dx = [ xn+1
n+1 ]1 0 = 1/(n + 1) as we conjectured, for any integer n ≥ 1.
(c) What does the answer to the second part imply about the average value of f(x) = x n
as n gets larger and larger? Can you explain this from the graphs of f(x) = x n ?
Solution: Therefore, as n → ∞, the average value of f(x) = x n on the interval [0, 1]
approaches 0 (for lim n→∞ 1/(n + 1) = 0). This makes sense, however, as we are considering
functions which take numbers x between 0 and 1 and raise them to high powers.
We think of 1/2 to larger and larger powers and see that it tends to 0 and any number
between 0 and 1 (except 1 itself) heads to 0 when raised to high enough powers.
8. For each of the following definite integrals, determine which technique — substitution or
integration by parts — will succeed in finding an antiderivative. [NOTE: You do not have
to find the integrals themselves, just do enough computation to be sure that a technique will
work.]
(a) ∫ x 3 ln xdx

Math 113, Calculus II
Test 1: Review Solutions
Solution: For this one, I’d use PARTS with u = ln x (so du = 1/x dx) and dv = x 3 dx
(so v = x 4 /4). Then
∫
x 3 ln xdx = 1 ∫ 1
4 x4 ln x −
1 4 x4 x dx = 1 4 x4 ln x − 1 ∫
4
x 3 dx = 1 4 x4 ln x − 1 x 4
4 4 + C.
(b) ∫ (ln x) 3
dx
x
Solution: For this one, let’s go with our first instinct and try the SUBSTITUTION
u = ln x. Then du = 1 xdx and we can write the integral as
∫ (ln x)
3
∫
dx =
x
u 3 du = u4
4 + C = 1 4 (ln x)4 + C.
(c) ∫ x(ln x) 3 dx
Solution: With this one, we’re back to trying to use PARTS with u = (ln x) 3 (so
du = 3(ln x) 2 1 x dx) and dv = xdx (so v = x2 /2) and
∫
x(ln x) 3 dx = 1 ∫ 1
2 x2 (ln x) 3 −
2 x2 · 3(ln x) 2 1 x dx = 1 2 x2 (ln x) 3 − 3 ∫
x(ln x) 2 dx
2
and we’d do the second integral using PARTS again in the same manner (with u = (ln x) 2
and dv = xdx) and so on, but by this point we can see that this method will work as the
integral ∫ x(ln x) 2 dx is simpler than ∫ x(ln x) 3 dx that we started with by decreasing the
exponent of ln x by one (and we would be correct to expect that the second application
will reduce the exponent of ln x by one more so we’d be left to integrate ∫ x ln xdx and
PARTS a third time would get rid of the ln x in the integral altogether.
(d) ∫ ln x
x 3 dx
Solution: We’re going to do PARTS once again, with u = ln x and dv = x −3 dx (so
du = 1/x dx and v = x −2 /(−2)) giving us
∫ ∫ ln x x−2 x
−2
dx =
x3 −2 ln x − 1
−2 x dx = − 1
2x 2 ln x + 1 ∫
2
x −3 dx = − 1
2x 2 ln x + 1 x −2
2 −2 + C.
(e) ∫ ln x 3
x
dx
Solution: For this one, we’ll use the SUBSTITUTION u = ln x after simplifying the
expression ln(x 3 ) = 3 ln x, so that du = 1 xdx and we have
(f) ∫ ln 3x
x
dx
∫ ln(x 3 ∫
)
x
dx =
3udu = 3 u2
2 + C = 3 2 (ln x)2 + C.
Solution: Here, we simplify ln 3x = ln 3 + ln x so that
∫ ∫
ln 3x
x
dx =
( ln 3
x
ln 3x
x = ln 3
x
+ ln x
x . Thus
+ ln x
x )dx = (ln 3)[ln |x|] + 1 2 (ln x)2 + C
using the SUBSTITUTION u = ln x and du = 1 xdx to compute the second integral.

Math 113, Calculus II
Test 1: Review Solutions
(g) ∫ ln x 3
dx
x 3
Solution: While substitution may look tempting, it isn’t what we want for this one.
Instead, after simplifying ln(x 3 ) = 3 ln x, we see that ∫ ln x 3
x 3 dx = ∫ 3 ln x
x 3 dx = 3 ∫ ln x
x 3 dx
is just 3 times the integral we computed in part (d) using PARTS (u = ln x and dv =
x −3 dx).
(h) ∫ (ln x) 3 dx
Solution: For this one, we have to use PARTS with u = (ln x) 3 (so du = 3(ln x) 2 1 x dx)
and dv = dx (so v = x). Then
∫
∫
(ln x) 3 dx = x(ln x) 3 −
x · 3(ln x) 2 1 ∫
x dx = x(ln x)3 − 3
(ln x) 2 dx
and we see that we’ve decreased the degree of ln x in the integral by one in the process
so the result of using parts is simpler than what we started with. Doing parts again on
∫
(ln x) 2 dx we would decrease the degree by one more, and another application of parts
gets rid of the ln x completely.
9. Let f be a continuous function with antiderivative F on the interval [a, b]. Let c be any point
in the interval. State whether the following are true or false. If false, then give an example
to show why it is false and correct the statement if possible If true, explain why.
(a) ∫ b
a f(x)dx = ∫ c
a f(x)dx = ∫ b
c f(x)dx
Solution: FALSE! If c = a, then ∫ c
a f(x)dx = 0 but certainly ∫ b
a
f(x)dx isn’t always
zero. The “mistake” was made by writing an equality sign rather than addition on the
right:
∫ b
a
f(x)dx =
∫ c
a
f(x)dx +
∫ b
c
f(x)dx.
(b) ∫ b
a
F (x)dx = f(b) − f(a)
Solution: FALSE! This is similar to the Evaluation Theorem, which says
∫ b
a
f(x)dx = F (b) − F (a),
but it makes the “mistake” of exchanging the roles of F and f.
(c) ∫ b
a f(x)dx ≥ 0
Solution: FALSE! If f(x) < 0 for all x in [a, b], say f(x) = −1, then ∫ b
a
f(x)dx will be
< 0 (in the example of f(x) = −1, we have ∫ b
a
−1dx = (−1)(b − a), which is negative if
b > a).
(d) ∫ b
a
cf(x)dx = c[F (b) − F (a)]
Solution: TRUE! First, ∫ b
then ∫ b
a
a cf(x)dx = c ∫ b
a
f(x)dx by the Constant Multiple Rule and
f(x)dx = F (b) − F (a) by the Evaluation Theorem.
(e) ∫ b
a
f(x)dx = f(m)(b − a) for some m in [a, b].
Solution: TRUE! This is the Mean Value Theorem for Integrals.
10. Let A(t) be the area under the curve y = sin(x 2 ) from 0 to t and B(t) be the area of the
triangle with vertices at O = (0, 0), P = (t, sin(t 2 )), and (t, 0). Find lim t→0 + A(t)/B(t).

Math 113, Calculus II
Test 1: Review Solutions
Solution: The value of B(t) is the area of the triangle, which is one-half its base (t) times
its height (sin(t 2 )), so B(t) = 1 2 t sin t2 while A(t) = ∫ t
0 sin x2 dx. To compute the limit, we’re
going to use l’Hôpital’s Rule several times and the Fundamental Theorem of Calclus in our
first application of it:
A(t)
lim
t→0 + B(t) = lim
1
2
∫ t
t→0 +
t→0 + ∫ t
= 2 lim
0 sin x2 dx
t sin t2
0 sin x2 dx
t sin t 2
= 2 lim
t→0 + sin t 2
sin t 2 + t cos(t 2 )[2t]
cos(t 2 )[2t]
= 2 lim
t→0 + cos(t 2 )[2t] + [4t] cos(t 2 ) + [2t 2 ][− sin(t 2 )[2t]]
= 2 lim
t→0 + [2] cos(t 2 ) + [2t][− sin(t 2 )[2t]]
[6] cos(t 2 ) + t(· · · )
= 2 [2] + 0
[6] + 0 = 2 3 .
11. Consider the function g(x) = ∫ 3
x
f(t)dt where the function f(t) is given by the graph on your
review sheet.
(a) On what intervals is g increasing? Decreasing?
Solution: The function g is increasing when its derivative g ′ (x) is positive and is decreasing
when g ′ (x) is negative. So we need to find g ′ (x), but g(x) = ∫ 3
x f(t)dt = − ∫ x
3 f(t)dt,
so the Fundamental Theorem of Calculus (Part I) implies that
g ′ (x) = d ∫ x
dx [− f(t)dt] = − d ∫ x
dx [ f(t)dt] = −f(x).
3
Hence, g is increasing when g ′ (x) = −f(x) > 0, i.e., when f(x) < 0, so g is increasing
on the interval (2, 4). Likewise, g is decreasing when g ′ (x) = −f(x) < 0, i.e., when
f(x) > 0, so g is decreasing on the intervals (0, 2) and (4, 6).
(b) At what values of x does g have a local max? Local min?
Solution: The function g has a local max where it changes from increasing to decreasing
(by the First Derivative Test), which occurs at x = 4. The function g has a local min
where it changes from decreasing to increasing, which happens at x = 2.
(c) On what intervals is g concave up? Concave down?
Solution: The function g is concave up when its second derivative g ′′ (x) = d
dx [g′ (x)] =
d
dx [−f(x)] = −f ′ (x) is positive, so g is concave up when f ′ (x) < 0. That is, g is concave
up when the graph of y = f(x) is decreasing, which is the case on the interval (1, 3).
The function g is concave down where the graph of f is increasing (so that f ′ (x) > 0),
which occurs on the intervals (0, 1) and (3, 5).
12. Show that the area enclosed by the graph of the parabola
3
y = f(x) = 2 a 2 x − 1 a 3 x2
(a > 0) and the x-axis has an area independent of a. How large is the area?

Math 113, Calculus II
Test 1: Review Solutions
∫ 2a
0
Solution: We first need to figure out what region we’re talking about, namely, when is
f(x) = 0? Clearly, f(0) = 0, but if we factor
2
a 2 x − 1 a 3 x2 = x (2a − x) = 0
a3 when x = 0 or x = 2a. Thus the area enclosed by the parabola is
( 2 a 2 x− 1 a 3 x2 )dx = [ 2 a 2 (x2 2 )− 1 a 3 (x3 3 )]2a 0 = [ x2
a 2 − x3
3a 3 ]2a
0 = [( (2a)2
a 2
− (2a)3 4a2
)−(0)] =
3a3 a 2 −8a3 3a 3 = 4−8 3 = 4 3 .
13. Consider the function f(x) = x 2 on the interval [0, 3], and the region R below the graph of f
on this interval. Find:
(a) The average value f ave on [0, 3] and the number x ave in [0, 3] at which it occurs (i.e.,
f(x ave ) = f ave ).
Solution: The average value of f(x) on [0, 3] is f ave = 3−0 ∫
3 0 x2 dx = 1 3 [ x3
3 ]3 0 = 1 3 [ 27 3 ] = 3.
The value of x ave is then the number in the interval [0, 3] such that f(x ave ), which is
x 2 ave, is equal to f ave = 3. Thus all we need do is solve the equation x 2 ave = 3 and select
the solution x ave = √ 3 = 1.732 that is in [0, 3].
(b) The value of x split in [0, 3] for which the vertical line x = x split divides R into two regions
of equal area.
Solution: To find x split that divides R into two regions of equal area, we first need to
know the total area of R, which is ∫ 3
0 x2 dx = [x 3 /3] 3 0 = 9. Thus we seek to find the
value of x split in [0, 3] so that ∫ x split
0
x 2 dx = 1 2 [9] = 9 2
. Now the value of our integral is
[x 3 /3] x split
0 = 1 3 x3 split , so we must solve 1 3 x3 split = 9 2 , which is done by x3 split = 27 2
, so that
x split = ( 27 2 )1/3 = 2.3811.
14. Calculate I(x) = ∫ e −x x 5 dx.
hint: This problem will be more fun if you forego repeated integration by parts and use The
Method of Undetermined Coefficients:
Make an educated guess that
I(x) = e −x ( a 0 + a 1 x + · · · + a 5 x 5) ,
Then compute I ′ (x), set it equal to the integrand e −x x 5 and determine the coefficients
a 0 . . . a 5 .
Solution: We find I ′ (x) = [−e −x ](a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 ) + (e −x )[a 1 + 2a 2 x +
3a 3 x 2 + 4a 4 x 3 + 5a 5 x 4 ], which must be equal to d
dx [∫ e −x x 5 dx] = e −x x 5 . We observe that we
can factor e −x out of both terms in our derived version of I ′ (x), so cancelling it with the e −x
in e −x x 5 , we find we must have −(a 0 +a 1 x+a 2 x 2 +a 3 x 3 +a 4 x 4 +a 5 x 5 )+(a 1 +2a 2 x+3a 3 x 2 +
4a 4 x 3 + 5a 5 x 4 ) = (a 1 − a 0 ) + (2a 2 − a 1 )x + (3a 3 − a 2 )x 2 + (4a 4 − a 3 )x 3 + (5a 5 − a 4 )x 4 − a 5 x 5 .
Equating coefficients, we have −a 5 = 1 so a 5 = −1; 5a 5 − a 4 = 0 so a 4 = −5; 4a 4 − a 3 = 0 so
a 3 = 4(−5) = −20; 3a 3 − a 2 = 0 so a 2 = 3(−20) = −60; 2a 2 − a 1 = 0 so a 1 = 2(−60) = −120;
and a 1 − a 0 = 0 so a 0 = −120. Hence
I(x) = e −x (−120 − 120x − 60x 2 − 20x 3 − 5x 4 − x 5 ).