Yahir Tucker

Answered

2022-06-24

We know if $g$ is continuous on $(a,b)$ and $F(x)={\int}_{a}^{x}g(t)dt$, then

${F}^{\prime}(x)=g(x)$

But, how about if we have

$F(x)={\int}_{a}^{h(x)}g(t)dt$

What should ${F}^{\prime}(x)$ be?? can we still apply fundamental theorem of calculus?

${F}^{\prime}(x)=g(x)$

But, how about if we have

$F(x)={\int}_{a}^{h(x)}g(t)dt$

What should ${F}^{\prime}(x)$ be?? can we still apply fundamental theorem of calculus?

Answer & Explanation

sleuteleni7

Expert

2022-06-25Added 28 answers

In that case you have :

${\int}_{a}^{h(x)}g(t)dt=G(h(x))-G(a)$

where ${G}^{\prime}(x)=g(x)$. Hence :

$[G(h(x)){]}^{\prime}={h}^{\prime}(x){G}^{\prime}(h(x))={h}^{\prime}(x)g(h(x))$

by the chain rule.

${\int}_{a}^{h(x)}g(t)dt=G(h(x))-G(a)$

where ${G}^{\prime}(x)=g(x)$. Hence :

$[G(h(x)){]}^{\prime}={h}^{\prime}(x){G}^{\prime}(h(x))={h}^{\prime}(x)g(h(x))$

by the chain rule.

telegrafyx

Expert

2022-06-26Added 8 answers

Yes, but you need also the chain rule. Obviously, $h$ must be differentiable.

${F}^{\prime}(x)={h}^{\prime}(x)g(h(x))$

${F}^{\prime}(x)={h}^{\prime}(x)g(h(x))$

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