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Evaluate the given improper integral

$$\int_0^\infty \int_0^\infty e^{-(2x+7y)}\mathop{\mathrm dy} \mathop{\mathrm dx} $$

Here is what I tried:

$$\int_0^\infty\lim\limits_{t \to \infty}[e^{-2x-7y}]_0^t \mathop{\mathrm dx}=\lim\limits_{t \to \infty} \int_0^t \frac{1}{7}e^{-2x}\mathop{\mathrm dx}= -\frac{1}{7}$$

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  • $\begingroup$ @Isko10986 $1/14$, I reckon. The integrand is always positive. $\endgroup$ – Fimpellizieri Nov 16 '16 at 4:38
  • $\begingroup$ @Fimpellizieri o yeah, I forgot to multiply the other negative. My bad. Should be $\frac{1}{14}$. Thanks! $\endgroup$ – Isko10986 Nov 16 '16 at 4:40
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Hint: this problem actually reduces to calculating

$$\left(\int_0^{\infty}e^{-2x}\,dx\right)\cdot\left(\int_0^{\infty}e^{-7y}\,dy\right)$$

Why?

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  • $\begingroup$ Thank you I got it. Is it because you can split up the x and y into two separate integrals? Here is the solution, I didn't want to manually edit because it takes long to type integrals and limits: imgur.com/a/dgz5d $\endgroup$ – asdfghjkl Nov 16 '16 at 4:51
  • $\begingroup$ That's right! Notice this happens because the exponential has the property that $$e^{a+b}=e^a \cdot e^b$$, so the variables can be separated. $\endgroup$ – Fimpellizieri Nov 16 '16 at 4:53

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