Question

How do you evaluate the integral from -1 to 1 of `int x^2/sqrt(x^3 +9) dx`?

Answer 1

The integral equals approximately `0.223`.

Explanation:

We have:

`I = int_-1^1 x^2/sqrt(x^3 + 9)dx`

Letting `u = x^3 + 9`, then `du = 3x^2dx` and `dx = (du)/(3x^2)`. We also need to readjust the bounds of integration, as such `u(1) = 1^3 + 9 = 10` and `u(-1) = (-1)^3 + 9 = 8`.

`I = int_8^10 x^2/sqrt(u) * (du)/(3x^2)`

`I = 1/3int_8^10 1/(u^(1/2))`

`I = 1/3int_8^10 u^(-1/2)`

`I = 1/3[2u^(1/2)]_8^10`

`I = 2/3sqrt(10) - 2/3sqrt(8)`

`I = 2/3(sqrt(10) - 2sqrt(2))`

`I ~~ 0.223`

Hopefully this helps!