Exponential improper integral

Problem

Compute $\displaystyle\int_0^{+\infty}e^{-2x}\,dx$.

Solution

It is an improper integral over an unbounded interval: we treat it as a limit. $\int_0^{+\infty}e^{-2x}\,dx = \lim_{t\to+\infty}\int_0^t e^{-2x}\,dx = \lim_{t\to+\infty}\left[-\frac{1}{2}e^{-2x}\right]_0^t = \lim_{t\to+\infty}\left(\frac{1}{2}-\frac{1}{2}e^{-2t}\right) = \boxed{\dfrac{1}{2}}.$ The term $e^{-2t}\to 0$, so the integral converges.

Links

Topics: Integral
Concepts: Improper integral · Limit
Skills: Compute limits · Integrate
Exercise types: Integral computation