Question 361330
*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = \int_{-\infty}^{0} e^{-(-x)}dx+\int_{0}^{\infty} e^{-(x)}dx] ... First break up the integral.



Note: We're using the idea that {{{f(x)=abs(x)}}} is a piecewise function and *[Tex \LARGE \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx] for some number c which is in the interval [a,b]



*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = \int_{-\infty}^{0} e^{x}dx+\int_{0}^{\infty} e^{-x}dx] ... Simplify



*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = \lim_{a\rightarrow -\infty}\int_{a}^{0} e^{x}dx+\lim_{b\rightarrow \infty}\int_{0}^{b} e^{-x}dx] ... Now introduce limits.



*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = \lim_{a\rightarrow -\infty}\left(\left. e^x \right|_{a}^{0}\right)+\lim_{b\rightarrow \infty}\left(\left. -e^{-x} \right|_{0}^{b}\right)] ... Take the antiderivative of {{{e^x}}} to get {{{e^x}}}. Take the antiderivative of {{{e^(-x)}}} to get {{{-e^(-x)}}}.



*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = \lim_{a\rightarrow -\infty}\left(\left. e^x \right|_{a}^{0}\right)-\lim_{b\rightarrow \infty}\left(\left. e^{-x} \right|_{0}^{b}\right)] ... Pull the negative out of the second limit.



*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = \lim_{a\rightarrow -\infty}\left(e^0-e^a\right)-\lim_{b\rightarrow \infty}\left(e^{-b}-e^0\right)] ... Evaluate at the limits of integration.



*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = \lim_{a\rightarrow -\infty}\left(1-e^a\right)-\lim_{b\rightarrow \infty}\left(e^{-b}-1\right)] ... Evaluate {{{e^0}}} to get 1.



*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = \left(1-0\right)-\left(0-1\right)] ... Evaluate each limit. 



*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = 1-(-1)] ... Subtract.



*[Tex \LARGE \int_{-\infty}^{\infty} e^{-\|x\|}dx = 2] ... Subtract.