Find a value of c> 1 so that the average value of f(x)=(9pi/x^2)cos(pi/x) on the interval [1, c] is -0.9.

Question

A. c=1
B.c=pi/4
C.c=6
D.c=4
E.c=pi/6

Please show the process

Answer 1

integrate the function from x = 1 to x = c and divide by (c-1)
I will say z = 1/x
then dx = -dz/z^2
when x = 1, z = 1
when x = c, z = 1/c
so
integral from z = 1 to z = 1/c
of -9 pi z^2 cos (pi z) (-dz/z^2)
= integral from z = 1 to z = 1/c
of 9 pi cos pi z dz
= 9 sin pi z
but z = 1/x
so 9 sin pi/x
at x = c - at x = 1
9 [ sin pi/c - sin pi] = 9 sin pi/c
divide by (c-1)
[9/(c-1)] sin pi/c = -.9
if c>1 the [9/(c-1)] is positive so the sin is negative
[1/(c-1)] sin pi/c = -.1
Does some c satisfy that ? I do not see it so must have an error, no surprise.

Answer 2

by the way c is not >1 for most of your answer choices, just C. and D.
Is there a typo somewhere?

Answer 3

you can search the question in chegg it will appear to you, I don't see the error in my typing

Answer 4

I have no relation to Chegg but could they not help you with it?

Answer 5

so, from sin(pi/c) = 1/10 (c-1)
the only solution I see right off with c>1 is c=6.
sin(pi/6) = 1/10 (6-1)
1/2 = 1/2