Answer
$$\frac{1}{3}$$
Work Step by Step
$$\eqalign{
& {\text{The area if the region is given by}} \cr
& A = \int_0^{ + \infty } {{e^{ - 3x}}} dx \cr
& {\text{Using the definition of improper integrals}} \cr
& A = \mathop {\lim }\limits_{b \to + \infty } \int_0^b {{e^{ - 3x}}} dx \cr
& A = \mathop {\lim }\limits_{b \to + \infty } \left[ { - \frac{1}{3}{e^{ - 3x}}} \right]_0^b \cr
& A = - \frac{1}{3}\mathop {\lim }\limits_{b \to + \infty } \left[ {{e^{ - 3x}}} \right]_0^b \cr
& A = - \frac{1}{3}\mathop {\lim }\limits_{b \to + \infty } \left[ {{e^{ - 3b}} - {e^0}} \right] \cr
& {\text{Evaluate the limit}} \cr
& A = - \frac{1}{3}\left[ {{e^{ - \infty }} - 1} \right] \cr
& A = - \frac{1}{3}\left[ {0 - 1} \right] \cr
& A = \frac{1}{3} \cr} $$
