To my understanding, the Leibniz Integral lets
$$\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt = \int_{a(x)}^{b(x)}\frac{d}{dx}f(x,t)dt$$
under certain restrictions. (Pretend the second $\frac{d}{dx}$ is a partial derivative, since I can't figure out how to do it).
I haven't seen the proof and the Wikipedia page is a bit dense for me, so I'm asking here. Can this rule also be applied in the case that $f(x,t) = f(t)$? My initial reaction makes me think "why not?", but if so, then any integral of that form would simply be $0$:
$$
\begin{align*}
\frac{d}{dx}\int_{a(x)}^{b(x)} f(t)dt &= \int_{a(x)}^{b(x)}\frac{d}{dx}f(t)dt\\
&= \int_{a(x)}^{b(x)}0dt\\
&= 0\Big|_{a(x)}^{b(x)}\\
&= 0
\end{align*}
$$
So simply, can the Leibniz integral be applied to integrals of this form? And if so, why is the result valid? And if not, what assumption does the proof rest on that makes it so?
Best Answer
What I think you are asking is: $$\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac d{dx}\left[F(b(x))-F(a(x))\right]=b'(x)F'(b(x))-a'(x)F'(a(x))$$ $$=b'(x)f(b(x))-a'(x)f(a(x))$$