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.

I've seen these types of questions before, but I think I missed a formal explanation of them. I have a solution to the question in front of me, so my question is not related to what the answer is - but what is the name for this type of problem? I don't quite understand why we're given these kinds of problems other than just because...

Find and simplify the derivative of $\displaystyle\int_{2/x}^{1}\frac{t^{3}}{1+t^{5}}\ dt$

\begin{align} f(x)&=\int_{2/x}^{1}\frac{t^{3}}{1+t^{5}}\ dt\\ &=-\int_1^{2/x}\frac{t^{3}}{1+t^{5}}\ dt\\ f'(x)&=\frac{(2/x)^3}{1+(2/x)^5}\cdot \left(-\frac{2}{x^2}\right)\\ &=\frac{16}{x^5+32} \end{align}

I'm trying to prepare for my final exam on Wednesday and I'd like to get a bit more background on this type of question so I can better understand what I'm doing if I am asked.

share|improve this question
    
I'm not sure if this is what you're after, but what you're using is the fundamental theorem of calculus. –  Git Gud Apr 14 '13 at 15:37
    
@GitGud I know what the FTC is... and I realize that it is being used in this process... but I'm curious about the type of problem where you're given a definite integral with variables in one or both of its limits. –  agent154 Apr 14 '13 at 15:40
    
I really don't understand what you're looking for. Hopefully someone else will. –  Git Gud Apr 14 '13 at 15:42
    
Check out Liebnitz rule –  Gautam Shenoy Apr 14 '13 at 15:45
add comment

1 Answer

up vote 2 down vote accepted

Let $f(t)$ be any integrable function and define $g(x)=\int_{a(x)}^{b(x)}f(t)dt$. Then, by the fundamental theorem of calculus, $g(x)=F(b(x))-F(a(x))$ where $F(x)$ is a function with $F'(t)=f(t)$. Now apply the chain rule: $$g'(x)=F'(b(x))\cdot b'(x)-F'(a(x))\cdot a'(x)=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x)$$ In your case, $b(x)=1$, so $b'(x)=0$ and $a(x)=2/x$ hence $a'(x)=-2/x^2$

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.