Expert: Paul Klarreich - 5/2/2009

Question
Let f(x) = (14 π)x^2 and g(x)= k^2sin(pi x/2k) for k>0

a. Find the average value of f on [1,4].
b. For what value of k will the average value of g on [0,k] be equal to the
average value of f on [1,4]?

So for
1. Integrate f from x=1 to x=4:
14 pi (4)^3/3 - 14pi (1)^3/3

then divide by the interval; i.e., 4-1 = 3

2. Integrate g from x=0 to x=k

-k^2/(pi/(2k)) * cos(pi (k)/(2k))
- [-k^2/(pi/(2k)) * cos 0]
note that cos 0 = 1

divide this by k-0 = k.

Equate this to the answer obtained in a and solve for k.
Is my approach correct?

Answer
Questioner:   Jeanne
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Mean Value theorm and average value
Question:  Let f(x) = (14 π)x^2 and g(x)= k^2 sin(pi x/2k) for k>0

a. Find the average value of f on [1,4].
b. For what value of k will the average value of g on [0,k] be equal to the
average value of f on [1,4]?

So for
1. Integrate f from x=1 to x=4:
14 pi (4)^3/3 - 14pi (1)^3/3

then divide by the interval; i.e., 4-1 = 3
..................
>> yes, that is correct.
..................

2. Integrate g from x=0 to x=k

-k^2/(pi/(2k)) * cos(pi (k)/(2k))
- [-k^2/(pi/(2k)) * cos 0]
note that cos 0 = 1

divide this by k-0 = k.
............
>> Looks right, too.
............


Equate this to the answer obtained in a and solve for k.
Is my approach correct?

>> Yes, indeed.  Let's check it out.

---------------------------------
(14 π)x^2 --> 14pi x^3/3 --> 14pi/3(64 - 1) = 14pi/3 (63) = 14(21)pi

divided by 3 --> 98 pi

-----------------------------

 
k^2 sin(pi x/2k) integrates to

- k^2 cos(pi x/2k) (2k)
------------------------
           pi

- 2k^3 cos(pi x/2k)
------------------------
           pi


Over [0,k], you have

- 2k^3 cos(pi k/2k)  - (- 2k^3 cos(0))
-----------------------------------------
              pi

- 2k^3 cos(pi /2)  - (- 2k^3 )
--------------------------------
              pi

- 2k^3 (0)  - (- 2k^3 )
--------------------------
              pi

2k^3
------
 pi

Whew! Sorry, I tend to get wordy here -- just trying to get it right.

Now you divide by k, because it is an average value:

2k^2
----
pi

finally  

2k^2
---- = 98 pi
pi


k^2 = 49 pi^2,

k = 7pi.

Did we get it right?

Note: This problem had nothing to do with the MVT. But I love MVT problems, so send them.