Improper Integrals — Question 6

Question 6 – Solution

Step 1: Use symmetry. Since $e^{-|x|}$ is even, $$I=2\int_0^\infty e^{-x}\,dx.$$ Step 2: Replace infinity with a limit. $$I=2\lim_{b\to\infty}\int_0^b e^{-x}\,dx.$$ Step 3: Integrate and evaluate. $$\begin{align*} I&=2\lim_{b\to\infty}[-e^{-x}]_0^b\\ &=2\lim_{b\to\infty}(1-e^{-b})\\ &=2(1-0)=2. \end{align*}$$ $$\boxed{I=2}$$