Question

How do you find the average value of `y=x^2sqrt(x^3+1)` on the interval `[0, 2]`?

Answer 1

The average value is `26/9`.

Explanation:

The formula for the average value of a function `A` is

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

where `F(x)` is continuous on `[a, b]`

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

Let `u = x^3 +1`. Then `du = 3x^2dx` and `dx = (du)/(3x^2)`.

`A = 1/2 int_1^9 x^2sqrt(u) (du)/(3x^2)`

`A = 1/6 int_1^9 sqrt(u) du`

`A = 1/6[2/3u^(3/2)]_1^9`

`A = 1/6[2/3(9)^(3/2) - 2/3(1)^(3/2)]`

`A = 1/6[18 - 2/3]`

`A = 3 - 1/9`

`A = 26/9`

Hopefully this helps!