Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Yes. To see why it works you need to check the proof of this result. –  Mhenni Benghorbal Oct 22 '13 at 0:12
3  
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.