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Is it correct to state that

$\frac{d}{ds} \int_{u=0}^{u=s} f(u)du = f(s)$

if $f(u)$ is continuous?

If so, what is the relevant theorem in action? If not, what else would be needed?

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The fundamental theorem of calculus is what you are looking for. –  Ross Millikan Oct 26 '12 at 4:18

1 Answer 1

up vote 5 down vote accepted

$\frac{d}{ds} \int_{0}^{s}f(u)du=\frac{d}{ds}(F(s)-F(0))=F'(s)=f(s)$ since $F(0)$ is a constant.

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