Differentiability of $\int_0^tx^2f(x)dt$ If $f(x)$ is continuous, how can I prove that $\int_0^tx^2f(x)dt$ is differentiable?
This is what I thought of:
Since $\int_0^tx^2f(x)dt=F(t)-F(0)$ for some function $F$ which is the antiderivative of $x^2f(x)$ so $F$ is differentiable. Thus $\int_0^tx^2f(x)dt$ is differentiable. 
Am I right?
 A: Assuming you mean $\int_0^tx^2f(x)\,dx$, then one form of the fundamental theorem of calculus says:

If $g$ is integrable on $[a,b]$, then $G(t)=\int_a^tg(x)\,dx$ is continuous on $[a,b]$ and differentiable at each point where $g$ is continuous.

As a consequence,

If $g$ is continuous on $[a,b]$, then $G(t)=\int_a^tg(x)\,dx$ is differentiable on $[a,b]$.

Applying this to your problem, take $g(x)=x^2 f(x)$. Then $g$ is continuous because it is a product of continuous functions. So $G$ is differentiable.
A: yes you are right. However this is not a formal proof.
Also, note that since $F$ is an antiderivative of $x^2f(x)$ then 
$$\frac{dF}{dt}=\frac{d}{dt}\Big(\int_0^tx^2f(x)dx\Big)=t^2f(t)$$
This is a specific form of the fundamental theorem of Integral Calculus. For a couple of different proofs see: 
http://alevel-ibhlsl-math.blogspot.gr/2014/01/theoretical-remarks-3.html
http://alevel-ibhlsl-math.blogspot.gr/2014/01/theoretical-remarks-4_12.html
and also the following might be useful: 
http://alevel-ibhlsl-math.blogspot.gr/2013/12/theoretical-remarks-2-some-insight-into.html
http://alevel-ibhlsl-math.blogspot.gr/2013/12/theoretical-remarks-1-indefinite.html
