Yahir Tucker

Answered

2022-06-24

We know if $g$ is continuous on $\left(a,b\right)$ and $F\left(x\right)={\int }_{a}^{x}g\left(t\right)dt$, then
${F}^{\prime }\left(x\right)=g\left(x\right)$
But, how about if we have
$F\left(x\right)={\int }_{a}^{h\left(x\right)}g\left(t\right)dt$
What should ${F}^{\prime }\left(x\right)$ be?? can we still apply fundamental theorem of calculus?

Answer & Explanation

sleuteleni7

Expert

2022-06-25Added 28 answers

In that case you have :
${\int }_{a}^{h\left(x\right)}g\left(t\right)dt=G\left(h\left(x\right)\right)-G\left(a\right)$
where ${G}^{\prime }\left(x\right)=g\left(x\right)$. Hence :
$\left[G\left(h\left(x\right)\right){\right]}^{\prime }={h}^{\prime }\left(x\right){G}^{\prime }\left(h\left(x\right)\right)={h}^{\prime }\left(x\right)g\left(h\left(x\right)\right)$
by the chain rule.

telegrafyx

Expert

2022-06-26Added 8 answers

Yes, but you need also the chain rule. Obviously, $h$ must be differentiable.
${F}^{\prime }\left(x\right)={h}^{\prime }\left(x\right)g\left(h\left(x\right)\right)$

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