Math 2250 Written HW #15 Solutions
1. True or False: If f (x) and g(x) are integrable functions on the interval [a, b], then
Z b
Z b
Z b
f (x)g(x) dx =
f (x) dx
g(x) dx .
a
a
a
If your answer is “true”, explain why. If your answer is “false”, give a counterexample.
Answer: False. We can get a counterexample by letting f (x) = x, g(x) = 2x, a = 0 and
b = 1 (in fact, almost every other choice you could make will also lead to a counterexample).
Then
Z 1
Z 1
Z b
2 3 1 2 3 2 3 2
2
2x dx =
x · 2x dx =
f (x)g(x) dx =
x
= (1) − (0) = .
3
3
3
3
0
0
a
0
On the other hand,
Z
b
b
Z
f (x) dx
a
a
Z 1
g(x) dx =
x dx
2x dx
0
0
!
2 1 2 1 x
x 0
=
2 0
2
1
02
=
−
12 − 02
2
2
1
= ,
2
Z
1
so we see that the statement is false.
2. Find the shaded area:
6
4
y ‡ x + sinHxL
2
Π
2Π
Answer: The shaded area is just the value of the definite integral from 0 to 2π of the function
f (x) = x + sin(x):
2π
2
Z 2π
x
(x + sin(x)) dx =
− cos(x)
2
0
0
2
0
(2π)2
=
− cos(2π) −
− cos(0)
2
2
2
4π
=
− 1 − (0 − 1)
2
= 2π 2 .
So the shaded area in the picture is exactly 2π 2 .
1
3. If the function f (x) is integrable on the interval [a, b], then the average value of f (x) on [a, b]
is defined to be
Z b
1
f (x) dx.
b−a a
(a) Show that this definition of average value gives the value you would expect for the
function f (x) = 2x on the interval [1, 3].
Answer: First of all, according to the definition given above, the average value of
f (x) = 2x is
Z 3
8
1
1 2 3 1 2
2x dx =
x 1=
3 − 12 = = 4.
3−1 1
2
2
2
On the other hand, the graph of the function f (x) = 2x on the interval [1, 3] is just the
straight line segment connecting the points (1, 2) and (3, 6), so the average value is just
the average of 2 and 6, which is 4. So indeed the average value defined above gives us
the answer we would expect in this case.
(b) The points (cos(θ), sin(θ)) as θ ranges from −π/2 to π/2 trace out a semicircular arc.
1
à
HcosHΘL, sinHΘLL
1
2
- 12
1
2
1
- 12
-1
In particular, notice that the function x(θ) = cos(θ) gives the x-coordinates of the points
on this semicircle. What is the average x-coordinate of a point on this semicircle? (Note:
the answer is not 1/2.)
Answer: Since the x-coordinates of points on the semicircle are just given by x(θ) =
cos(θ) and since the possible values of θ range from −π/2 to π/2, the average x-coordinate
is just the average value of the function cos(θ) on the interval [−π/2, π/2]. By the
definition above, this is
Z
π/2 − (−π/2) π/2
1 π/2
R
cos(θ) dθ =
cos(θ) dθ
π −π/2
−π/2
iπ/2
1h
=
sin(θ)
π
−π/2
1
= [sin(π/2) − sin(−π/2)]
π
1
= [1 − (−1)]
π
2
= .
π
2
So the average x-coordinate of a point on the semicircle is
3
2
π
≈ 0.63662.