= Differentiation of parameter integrals on compact rectangles
{title2=$\dfrac{d}{dt}\int_a^bf(t,x)\,dx=\int_a^b\partial_t f(t,x)\,dx$}
Suppose $f$ and its <partial derivative> $\partial_t f$ are jointly continuous on a compact rectangle $J\times[a,b]$. Then $F(t)=\int_a^bf(t,x)\,dx$ is continuously differentiable in the interior of $J$, and the displayed formula holds, with one-sided <derivatives> at parameter endpoints where the corresponding partial derivative is given.
To prove it, the <extreme value theorem> bounds $|\partial_t f|$ by $M_1$. For each fixed $x$, the <mean value theorem> bounds the difference quotient $[f(t+h,x)-f(t,x)]/h$ by $M_1$. It converges pointwise to $\partial_t f(t,x)$, and the constant $M_1$ is integrable over $[a,b]$. <Dominated convergence> gives the derivative formula. Apply <continuity of parameter integrals on compact rectangles> to $\partial_t f$ to obtain continuity of $F'$. The proof works locally on unbounded parameter <intervals> as well.
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