Question

What is the average value of the function `f(x)=(x^2 +4)/(x^2)` on the interval `[3,5]`?

Answer 1

`19/15`

Explanation:

The average value of the function `f` on the interval `[a,b]` is equal to:

`1/(b-a)int_a^bf(x)dx`

Here this gives us an average value of:

`1/(5-3)int_3^5(x^2+4)/x^2dx=1/2int_3^5(1+4x^-2)dx`

Using `intdx=x` and `intx^ndx=x^(n+1)/(n+1)` for `n!=-1`:

`=1/2[x-4/x]_3^5=1/2(5-4/5)-1/2(3-4/3)=19/15`