Question

If the average value of the function `v(t)=6/x^2` on the interval [1,c] is equal to 1, how do you find the value of c?

Answer 1

`c=6`

Explanation:

The average value of the function `f(x)` on the interval `[a,b]` is

`"average value"=1/(b-a)int_a^bf(x)dx`

With the given information, this gives us

`1=1/(c-1)int_1^c6/x^2dx`

This can be simplified as

`1=6/(c-1)int_1^cx^-2dx`

The integration here gives `intx^-2dx=-x^-1+C`, so we will evaluate the integral using this function:

`1=6/(c-1)[-1/x]_1^c`

Evaluating, this gives:

`1=6/(c-1)[-1/c-(-1/1)]`

`1=6/(c-1)(1-1/c)`

`1=6/(c-1)((c-1)/c)`

`1=(6c-6)/(c^2-c)`

`c^2-c=6c-6`

`c^2-7c+6=0`

`(c-6)(c-1)=0`

`c=1,6`

The solution `c=1` is thrown out because it causes `6/(c-1)` to have an undefined value. Thus, the only valid solution is `c=6`.