WORKSHEET FOR 1/23/2009 - SOLUTIONS
(1) Suppose that Rx
0f(t)dt = 3x2+ex−cos x. Find f(2).
Solution: We must use a version of the fundamental theorem of calculus, that states
that: d
dx Zx
a
f(t)dt =f(x).
With this in mind, we can differentiate both sides of the equation given in the problem
to get:
d
dx Zx
0
f(t)dt =d
dx(3x2+ex−cos x)
f(x) = d
dx(3x2+ex−cos x) (by fundamental theorem of calculus)
f(x) = 6x2+ex+ sin x
f(2) = 24 + e2+ sin 2.
(2) A valve is opened, and water starts flowing through a pipe at time t= 0. Suppose
that f(t) gallons/second of water are flowing through a pipe for any time t≥0.
Write down an expression for the amount of water than flows through the pipe in its
first 30 seconds of operation.
Solution: Z30
0
f(t)dt.
(3) Let F(x) = Rx
0(cos t)2etdt.
(a) Find the critical numbers of F.
(b) Where is Fconcave up and concave down? Find local maxima and minima on
the interval [−1,2π]?
(c) Find the where Fachieves its maximum and minimum values on the interval
[−1,2π].
Solution:
(a) By the fundamental theorem of calculus:
F0(x) = d
dx Zx
0
(cos t)2etdt = (cos x)2ex
Since ex6= 0, this function has critical numbers whenever cos x= 0, i.e. for
x=π
2+kπ, where kis an integer.
(b) Omitted.
(c) Since (cos t)2et≥0, cumulative area must be increasing. Thus F(x) never
decreases, so the minimum must occur at the left enpoint −1, and the maximum
must occur at the right endpoint 2π.
(4) Show that 0 ≤Rπ
0sin2xdx ≤π.Hint. You will not be able to compute the actual
area. Draw a graph and think about the area.
Solution: Since 0 ≤sin2x≤1, we have that Rπ
00dx ≤Rπ
0sin2xdx ≤Rπ
01dx, and
1
