TITLE: evaluating using fundamental theorem of calculus
QUESTION [7 upvotes]: Let  $$F(x)=\int_0^{x}\ tf  (x^2-t^2)\,dt$$
Find $F'(x)$. 
I know that I need to apply the fundamental theorem of calculus. As for the next part, I tried to substitute in $u=x^2-t^2$ but I don't know how to proceed.

REPLY [8 votes]: If you substitute $u = x^2 - t^2$ then $du = -2t \, dt$. If $t = 0$ then $u = x^2$. If $t = x$ then $u = 0$. Thus, we have
$$ F(x) = -\frac{1}{2} \int_{0}^x f(x^2 - t^2) (-2t) \, dt = -\frac{1}{2} \int_{x^2}^{0} f(u) \, du = \frac{1}{2} \int_0^{x^2} f(u) \, du. $$
Using the fundamental theorem of calculus, you have
$$ F'(x) = \frac{1}{2} f(x^2) (x^2)' = x f(x^2). $$

REPLY [2 votes]: Hint:
We have
$$
F(x) = \int_{0}^{x}tf(x^{2}-t^{2})dt = \frac{1}{2}\int_{0}^{x^{2}}f(u)du
$$
for all suitable $x$; here we used the substitution $u := x^{2}-t^{2}$. Now $F$ is a composite map; applying chain rule along with the fundamental theorem to conclude.