Question

What is the average value of the function `h (x) = 3/(1 + x)^2` on the interval `[0,2]`?

Answer 1

The average value is `1`.

Explanation:

The following formula represents the average value of a function `f(x)` in the interval [a, b]:

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

We have `b = 2` and `a = 0`. Our function to integrate is `3/(1 + x)^2`

`=1/(2 - 0) int_0^2 3/(1+ x)^2dx`

`=1/2int_0^2 3/(1 + x)^2dx`

It is true we could use partial fractions, but for this problem it would be simpler to use substitution. Let `u = x+ 1`. Then `du = dx`.

`=1/2int_0^2 3/u^2du`

`=1/2int_0^2 3u^-2du`

Integrate using `intx^ndx= x^(n + 1)/(n + 1) + C`, where `n != -1`.

`=1/2[-3/u]_0^2`

`=1/2[-3/(x + 1)]_0^2`

Evaluate using the second fundamental theorem of calculus, which states that `int_a^bF(x) = f(b) - f(a)`, where `f(x)` is continuous and differentiable on `[a, b]` and `f'(x) = F(x)`.

`=1/2(-3/3 - (-3/1))`

`=1/2(-1 + 3)`

`=1`

Hopefully this helps!