Answer
$a = \ln 3$
Work Step by Step
$$\eqalign{
& {\text{The area of the region is given by}} \cr
& A = \int_{ - a}^a {{e^{ - x}}} dx \cr
& {\text{Where }}A = \frac{8}{3},{\text{ therefore}} \cr
& \int_{ - a}^a {{e^{ - x}}} dx = \frac{8}{3} \cr
& {\text{Integrating}} \cr
& \left[ { - {e^{ - x}}} \right]_{ - a}^a = \frac{8}{3} \cr
& - \left[ {{e^{ - a}} - {e^{ - \left( { - a} \right)}}} \right] = \frac{8}{3} \cr
& - \left[ {{e^{ - a}} - {e^a}} \right] = \frac{8}{3} \cr
& {e^a} - {e^{ - a}} = \frac{8}{3} \cr
& {\text{Multiplying both sides of the equation by 3}}{e^a} \cr
& 3{e^a}\left( {{e^a} - {e^{ - a}}} \right) = 3{e^a}\left( {\frac{8}{3}} \right) \cr
& 3{e^{2a}} - 3{e^{a - a}} = 8{e^a} \cr
& 3{e^{2a}} - 3 = 8{e^a} \cr
& 3{e^{2a}} - 8{e^a} - 3 = 0 \cr
& {\text{Let }}x = {e^a} \cr
& 3{x^2} - 8x - 3 = 0 \cr
& {\text{Factoring}} \cr
& \left( {x - 3} \right)\left( {3x + 1} \right) = 0 \cr
& x = 3,{\text{ }}x = - \frac{1}{3} \cr
& {\text{Write in terms of }}{e^a} \cr
& {e^a} = 3,{\text{ }}\underbrace {{\text{ }}{e^a} = - \frac{1}{3}}_{{\text{No real solution}}} \cr
& {e^a} = 3 \cr
& {\text{Solve for }}a \cr
& \ln {e^a} = \ln 3 \cr
& a = \ln 3 \cr} $$
