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Consider the following, where $a(x)$, and $b(x)$ are both functions that are continuous in their domain:

$$ g(x) = \int\limits_{a(x)}^{b(x)}f(t)dt $$

Is it the case that $g'(x)$ is always the following?

$$ g'(x) = f(b(x))\times b'(x) - f(a(x))\times a'(x) $$

If so, why does this work?

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Yes. To see why it works you need to check the proof of this result. – Mhenni Benghorbal Oct 22 '13 at 0:12
Your notation here is incorrect. You can't use the same variable as both a free variable and a dummy variable of integration. Should be something like: $g(x) = \int\limits_{a(x)}^{b(x)}f(u)du$ – David H Oct 22 '13 at 0:14
Ah thanks david. You are correct, I will fix that. – Scuba Steve Oct 22 '13 at 1:28
up vote 5 down vote accepted

Fix a number $x_0\in\mathbb{R}$. You can then write $$g(x)=\int_{a(x)}^{b(x)}f(t)dt=\int_{x_0}^{b(x)}f(t)dt-\int_{x_0}^{a(x)}f(t)dt$$

Now, we want to write $\int_{x_0}^{b(x)}f(t)dt$ as a composition of "known" functions, and similarly with $\int_{x_0}^{a(x)}f(t)dt$.

Let $h(x)=\int_{x_0}^x f(t)dt$. By the fundamental theorem of calculus, $h'=f$. Now, notice that $$g=h\circ b-h\circ a.$$ By the chain rule, $$g'(x)=h'(b(x))\cdot b'(x)-h'(a(x))\cdot a'(x)=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x)$$ as wanted.

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