Download 6.2 Setting Up Integrals: Volume, Density, Average Value

720
CHAPTER 6
APPLICATIONS OF THE INTEGRAL
SOLUTION Let f .x/ D 2 sin x " x. A graph of y D f .x/ is shown below. From this graph, the positive root of f .x/ appears to
be roughly x D 1:9.
y
0.8
0.6
0.4
0.2
x
0.5
1.0
1.5
2.0
(a) Using a computer algebra system, solving the equation
2 sin ˛ " ˛ D 0
yields ˛ D 1:895494267.
(b) The average value of f .x/ on Œ0; ˛! is
M D
1
˛"0
Z
˛
0
f .x/ dx D 0:4439980667:
(c) Solving
f .c/ D 2 sin c " c D 0:4439980667
yields either c D 0:4805683082 or c D 1:555776337.
59. Which of f .x/ D x sin2 x and g.x/ D x 2 sin2 x has a larger average value over Œ0; 1!? Over Œ1; 2!?
SOLUTION The functions f and g differ only in the power of x multiplying sin2 x. It is also important to note that sin2 x ! 0
for all x. Now, for each x 2 .0; 1/, x > x 2 so
f .x/ D x sin2 x > x 2 sin2 x D g.x/:
Thus, over Œ0; 1!, f .x/ will have a larger average value than g.x/. On the other hand, for each x 2 .1; 2/, x 2 > x, so
g.x/ D x 2 sin2 x > x sin2 x D f .x/:
Thus, over Œ1; 2!, g.x/ will have the larger average value.
60. Find the average of f .x/ D ax C b over the interval Œ"M; M !, where a, b, and M are arbitrary constants.
SOLUTION
The average is
1
M " ."M /
61.
negative.
SOLUTION
Z
M
!M
.ax C b/ dx D
1
2M
Z
M
!M
.ax C b/ dx D
$ˇˇM
1 #a 2
x C bx ˇˇ
D b:
2M 2
!M
Sketch the graph of a function f .x/ such that f .x/ ! 0 on Œ0; 1! and f .x/ % 0 on Œ1; 2!, whose average on Œ0; 2! is
Many solutions will exist. One could be
y
1
1
2
x
−1
−2
62. Give an example of a function (necessarily discontinuous) that does not satisfy the conclusion of the MVT for Integrals.
SOLUTION There are an infinite number of discontinuous functions that do not satisfy the conclusion of the Mean Value Theorem
for Integrals. Consider the function on Œ"1; 1! such that for x < 0, f .x/ D "1 and for x ! 0, f .x/ D 1. Clearly the average value
is 0 but f .c/ ¤ 0 for all c in Œ"1; 1!.