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

Question

A. 3/2 B. 3 C. 6 D. 12 E.24

I've used u-sub and I got u=x^2-1 du=2x but I don't really know how to continue after that or integrate the u sub into the f(u)du=6 part

Answer 1

U = X^2 - 1

dU = 2x dx

integral x f ( x^2 - 1) dx =

1/2 * integeral 2 x f(x^2 - 1) dx =

1/2 * integral f(u) dU

for x=1, u = 1^2-1 =0, so the LOI are correct

1/2{ integral f(u) du } = 1/2(6) = 3