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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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    You could use the fact that $\int_{0}^{\infty} x^{k}e^{-cx} = \frac{k!}{c^{k+1}}$.2010-10-27

2 Answers 2

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

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    @Rasmus: We $c$an't but then the question he asked was by subsitution2010-10-06
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If you expand the brackets and write the integral term-by-term, then we get \begin{equation*} 4\int e^{-6x}dx+16\int e^{-10x}dx+16\int e^{-14x}dx \\ =[-\frac{8}{7}e^{-14x}-\frac{8e^{-10x}}{5}-\frac{2e^{-6x}}{3}]^{\infty}_{0}. \end{equation*}