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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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    That's a perfectly fine approach, and it's quite general too.2010-10-06
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    That's the approach I would use.2010-10-06
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    OK, thanks for the replies :)2010-10-06
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    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.2010-10-06
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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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    And how would you solve *this* other than by expanding?2010-10-06
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    @Rasmus: We can'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*}