Confused on the final step and want to double check my work.
Say $\int_9^{16} f(x)dx = 64$, what is value of $\int_9^{16} f(x^2)xdx$
I used u substitution and set $u = x^2$, then $\frac{du}{2} = xdx$
Changed the bounds since $u=x^2$, take the square root of each.
Then $\frac{1}{2}\int_3^{4} f(u)du =$ ?
No, you have not changed your limits of integration from $x$ to $u$.
Notice that if $$x=9, u=x^2=81$$
We simply do not have enough information to find $$ \int_9^{16} f(x^2)xdx$$