A simple Integral Question

Question

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

Answer 1

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 $c=6$.

Answer 2

The nice thing here is that you can rewrite this as \begin{equation} 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. \end{equation} Each of the integrals involved is simple to evaluate, and from there, $c$ falls right out.