Answer
$A = \sqrt 2 {e^{ - \frac{1}{2}}}$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can see that the area of the rectangle }} \cr
& {\text{inscribed under the curve }}f\left( x \right) = {e^{ - {x^2}}}{\text{given by:}} \cr
& A = 2xy \cr
& {\text{Where }}y = f\left( x \right) = {e^{ - {x^2}}},{\text{ then}} \cr
& A = 2x\left( {{e^{ - {x^2}}}} \right) \cr
& A = 2x{e^{ - {x^2}}} \cr
& {\text{Differentiating}} \cr
& \frac{{dA}}{{dx}} = 2x\left( { - 2x{e^{ - {x^2}}}} \right) + {e^{ - {x^2}}}\left( 2 \right) \cr
& \frac{{dA}}{{dx}} = 2{e^{ - {x^2}}} - 4{x^2}{e^{ - {x^2}}} \cr
& {\text{Let }}\frac{{dA}}{{dx}} = 0 \cr
& 2{e^{ - {x^2}}} - 4{x^2}{e^{ - {x^2}}} = 0 \cr
& 2{e^{ - {x^2}}}\left( {1 - 2{x^2}} \right) = 0 \cr
& 2{e^{ - {x^2}}} = 0,{\text{ or }}1 - 2{x^2} = 0 \cr
& 2{e^{ - {x^2}}} > 0{\text{ for all }}x{\text{ real number}}{\text{, then}} \cr
& 1 - 2{x^2} = 0 \cr
& 2{x^2} = 1 \cr
& x = \pm \sqrt {\frac{1}{2}} \cr
& x = \pm \frac{{\sqrt 2 }}{2} \cr
& {\text{Taking }}x > 0 \cr
& x = \frac{{\sqrt 2 }}{2} \cr
& {\text{Finding }}y{\text{ at }}x = \frac{{\sqrt 2 }}{2} \cr
& f\left( {\frac{{\sqrt 2 }}{2}} \right) = {e^{ - {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}}} \cr
& f\left( {\frac{{\sqrt 2 }}{2}} \right) = {e^{ - \frac{1}{2}}} \cr
& {\text{The area is given by}} \cr
& A = 2xy \cr
& A = 2\left( {\frac{{\sqrt 2 }}{2}} \right){e^{ - \frac{1}{2}}} \cr
& A = \sqrt 2 {e^{ - \frac{1}{2}}} \cr} $$
