Question

The average value of a function v(x)=2/x^2 on the interval [1,c] is equal to 1. How to find the value of c?

Answer 1

The answer is `c = 2`

Explanation:

Average value is given by

`A = 1/(b - a) int_a^b v(x)dx`

`1 = 1/(c - 1) int_1^c 2/x^2 dx`

`1 = 1/(c- 1) [-2/x]_1^c`

`c - 1 = -2/c - (-2/1)`

`c -1 = -2/c + 2`

`c - 3 = -2/c`

`c^2 - 3c = -2`

`c^2 - 3c + 2 = 0`

`(c - 2)(c - 1) = 0`

`c = 2 or 1`

But `c != 1`, since it would make `1/(1 - 1) = 1/0` which is impossible.

Therefore `c = 2`.

Hopefully this helps!