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

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...

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That's a perfectly fine approach, and it's quite general too. – Qiaochu Yuan Oct 6 '10 at 19:19
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
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

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 $$-\int\limits_{1}^{0} \Bigl[2t^{3}+4t^{7}\Bigr]^{2} \cdot \frac{dt}{t}$$