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Suppose $f$ and its partial derivative ${\partial }_{t}f$ are jointly continuous on a compact rectangle $J\times [a,b]$. Then $F(t)={\int }_a^{b}f(t,x)dx$ 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 $F^{\prime }$. The proof works locally on unbounded parameter intervals as well.

Fix $t_0\ge 0$. On a compact parameter interval containing all nonnegative $t$ sufficiently close to $t_0$, the corresponding rectangle with $x\in [0,1]$ is compact. The continuous function $f$ is therefore bounded there by some $M<\infty$, by the extreme value theorem.

For any sequence $t_n\ge 0$ tending to $t_0$, all sufficiently late $t_n$ lie in this rectangle's parameter range. For each $x\in [0,1]$, continuity givesThe constant function $M$ is integrable on $[0,1]$. The dominated convergence theorem therefore givesThe sequential criterion for continuity on a real interval proves $F$ is continuous on $[0,\infty )$. At zero this is right-continuity. This is continuity of parameter integrals on compact rectangles; the bound is local in the parameter, so $f$ need not be bounded on the entire unbounded strip.