Given an intergral $$ \int_{a(t)}^{b(t)} f(t, x) dx $$ What is the fomula to of its derivation? $$ \frac{\partial}{\partial t} \int_{a(t)}^{b(t)} f(t, x) dx $$

Would someone please give me a link to the fomula and its derivations?


The answer is $$ \int_{a(t)}^{b(t)} f'_t(t, x)dx + f(t, b(t)) b'_t(t) - f(t, a(t)) a'_t(t) $$ and here(Leibniz_integral_rule) is the reference.

Many thanks to André Nicolas's comment!

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    $\begingroup$ This link to the Leibniz Integral Rule may help. You can also search under that name, or under the name differentiation under the integral sign. $\endgroup$ Nov 19, 2013 at 8:35
  • $\begingroup$ @Integrator: Ordinarily when a simple link is involved, with no "added value" I prefer to just leave a comment, but I ca see your point, the site explicitly discourages that. $\endgroup$ Dec 15, 2014 at 2:58

1 Answer 1


By the Leibniz Rule for differentiation under the integral sign, we have $$\frac{d}{dt}\int_{a(t)}^{b(t)} f(t,x)\,dx =b'(t)f(t,b(t))-a'(t)f(t,a(t))+ \int_{a(t)}^{b(t)} \frac{\partial}{\partial t}f(t,x)\,dx .$$ For details, please see the Wikipedia article linked to above.


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