Answer
$$A = - {e^{1/3}} + e$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{{x^2}}}{e^{1/x}},{\text{ }}y = 0,{\text{ }}1 \leqslant x \leqslant 3 \cr
& {\text{From the graph we can define the area as:}} \cr
& A = \int_1^3 {\left( {\frac{1}{{{x^2}}}{e^{1/x}}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ { - {e^{1/x}}} \right]_1^3 \cr
& A = - {e^{1/3}} + {e^{1/1}} \cr
& A = - {e^{1/3}} + e \cr
& A \approx 1.322 \cr} $$
