Solutions to Practice Problems

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