The function f is continuous and ∫0 to 8 f(u)du=6. What is the value ∫ 1 to xf(x^2-1)dx? Simplify completely. Please help! Thank you.

Question

The function f is continuous and ∫0 to 8 f(u)du=6. What is the value ∫ 1 to xf(x^2-1)dx? Simplify completely. Please help! Thank you.

Answer 1

First, I'm going to assume that you left the upper bound of the integral you are trying to find off, otherwise the problem doesn't make sense. So what we should be finding is I = ∫₁³ x*f(x^2 - 1) dx.

To solve this, we want to use u-substitution, where u = x^2-1. Now du = 2x dx, so after u-substitution our integral I = ∫₀⁸ 1/2 f(u) du = 6/2 = 3. The second equals comes from the information we are already given. Also, the bounds are gotten by plugging in the bounds 1 and 3 for x into u (i.e. u = 1-1 = 0 and u = 9-1=8).