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How would one reasonable differentiate the integral of the form $\int_{f(x)}^{g(x)} h(x) dx$, provided $h$ is continuous? Does this even make reasonable sense? The main point is that the integrand is also a function of $x$.

mathgenesis22813
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    This has been asked many times here, see: https://en.wikipedia.org/wiki/Leibniz_integral_rule (see Example 2) – Moo Jun 23 '18 at 15:07

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$$\int_{f(x)}^{g(x)} h(x) dx = \int_{0}^{g(x)} h(x) dx -\int_{0}^{f(x)} h(x) dx$$

Apply the Fundamental Theorem of Calculus to get

$$ \frac {d}{dx} \int_{f(x)}^{g(x)} h(x) dx = h(g(x))g'(x) - h(f(x))f'(x) $$

Mohammad Riazi-Kermani
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