Answer
$$A = e$$
Work Step by Step
$$\eqalign{
& y = {e^x},{\text{ }} - \infty < x \leqslant 1 \cr
& {\text{The area is given by}} \cr
& A = \int_{ - \infty }^1 {{e^x}} dx \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_{ - \infty }^b {f\left( x \right)dx = \mathop {\lim }\limits_{a \to - \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr
& A = \mathop {\lim }\limits_{a \to - \infty } \int_a^1 {{e^x}} dx \cr
& {\text{Integrate}} \cr
& A = \mathop {\lim }\limits_{a \to - \infty } \left[ {{e^x}} \right]_a^1 \cr
& A = \mathop {\lim }\limits_{a \to - \infty } \left[ {{e^1} - {e^a}} \right] \cr
& A = \mathop {\lim }\limits_{a \to - \infty } \left( {e - {e^a}} \right) \cr
& {\text{Evaluate }}a \to - \infty \cr
& A = e - {e^{ - \infty }} \cr
& A = e - 0 \cr
& A = e \cr} $$
