Answers to Review Sheet for Final Exam

70. For the even function f graphed in the figure:
Let A = 2
−2 f(x) dx, B = 5
0 f(x) dx, C = 5
f(x) dx, D =
2
5
−2 f(x) dx, E = 0
−2 f(x) dx, and F = 2
f(x) dx.
0
(a) Write C in terms of A and B.
Solution: First of all, we know A = E + F and
B = F + C no matter what since we can break up
the integrals by adding an extra endpoint. Secondly,
by evenness we know that E = F . So A = 2F and
hence F = A/2. Thus solving B = F + C for C, we
get C = B − F = B − A/2.
(b) Write C in terms of D and E.
Solution: D = E + F + C and F = E by evenness,
so C = D − 2E.
(c) Write F in terms of C and D.
Solution: Again D = E + F + C and F = E, so
F = (D − C)/2.
71. Suppose you know that with n subdivisions, the right-hand sum for 4
f(x) dx is given by the
1
formula
3n + 5
Rn = .
n
(a) Suppose f(1) = 3 and f(4) = 5. What is the left-hand sum Ln?
Solution: Since
[f(4) − f(1)] · [4 − 1]
Rn − Ln = =
n
6
n ,
we know Ln = Rn − 6 3n−1 = n n .
(b) What is 4
f(x) dx?
1
Solution: It’s the limit of either the left or right sum as n → ∞, which is 3.
72. The area under 1/ √ x on the interval 1 ≤ x ≤ b is equal to 6. Find the value of b using the
Fundamental Theorem.
Solution:
b
1
x −1/2 dx = 2 √ b − 2 = 6. So b = 16.
73. Find the exact area of the region bounded by the x-axis and the graph of y = x 3 − x.
Solution: The cubic crosses the axis at three points: x = 0, x = 1, and x = −1. There is
both a region above the axis (for −1 < x < 0) and a region below the axis (for 0 < x < 1),
and we want them both. By symmetry they will have the same area. So if we just compute
one, we’ll know both.
1
0
(x 3 − x) dx =
1
4 x4 − 1
2 x2
x=1
= −
x=0
1
4 ,
which is negative as we expect since this region is below the axis. Therefore the area of each
region is 1
1
, and the total area is 4 2 .
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