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'm currently teaching myself calculus and I'm probably trying to run before I can walk, but I've been working on this problem..

I managed to find the correct result for:

$$\int_{0}^{\infty }(2e^{-3x}+4e^{-7x})^2dx$$

by expanding it to:

$$\int_{0}^{\infty}4e^{-6x}+16e^{-10x}+16e^{-14x}dx$$

and then working from there.

Is there a better/more general approach I could have taken? I've attempted to solve it using substitution but haven't had any success...

share|improve this question
12  
That's a perfectly fine approach, and it's quite general too. –  Qiaochu Yuan Oct 6 '10 at 19:19
3  
That's the approach I would use. –  You Oct 6 '10 at 19:22
    
OK, thanks for the replies :) –  mash Oct 6 '10 at 19:30
2  
I agree with the others, what you did is the best thing. In contrast to calculate the derivative of an explicit elementary function there is no general recipe to calculate the integral of an elementary function - instead we are forced to simplify until we reach something we know, as you did. –  AD. Oct 6 '10 at 21:35
    
You could use the fact that $\int_{0}^{\infty} x^{k}e^{-cx} = \frac{k!}{c^{k+1}}$. –  PEV Oct 27 '10 at 4:56
add comment

1 Answer 1

Using substitution you can actually solve this, but generally it's more or less the same thing. Put $t=e^{-x}$ therefore you have $dx = -\frac{1}{t} \ dt$. Then your integral reduces to $$-\int\limits_{1}^{0} \Bigl[2t^{3}+4t^{7}\Bigr]^{2} \cdot \frac{dt}{t}$$

share|improve this answer
    
And how would you solve this other than by expanding? –  Rasmus Oct 6 '10 at 20:32
    
@Rasmus: We can't but then the question he asked was by subsitution –  anonymous Oct 6 '10 at 20:42
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.