# A simple Integral Question

What are the steps to calculate the value of c in the following integral equation? $$\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}c.e^{-(x_1+2x_2+3x_3)}\,dx_1 \, dx_2 \, dx_3 = 1$$

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## 2 Answers

Since $e^{-x_1+2x_2+3x_3}=e^{-x_1}e^{-2x_2}e^{-3x_3}$, we have $$\int_0^\infty\int_0^\infty\int_0^\infty e^{-x_1+2x_2+3x_3}=\left(\int_0^\infty e^{-x_1}\right)\left(\int_0^\infty e^{-2x_2}\right)\left(\int_0^\infty e^{-3x_3}\right)=1\,\frac12\,\frac13=\frac16.$$

So $c=6$.

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The nice thing here is that you can rewrite this as $$c\int_0^\infty e^{-3x_3}\int_0^\infty e^{-2x_2}\int_0^\infty e^{-x_1}\,dx_1\,dx_2\,dx_3=1.$$ Each of the integrals involved is simple to evaluate, and from there, $c$ falls right out.

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