Answer
$\left( {\overline x ,\overline y } \right) = \left( {\frac{2}{{{e^2} - 1}},\frac{{{e^2} + 1}}{{4e}}} \right)$
Work Step by Step
$$\eqalign{
& {\text{From the given graph}}{\text{, we can estimate the centroid at}} \cr
& \left( {0.3,0.5} \right) \cr
& \cr
& y = {e^x},{\text{ }} - 1 \leqslant x \leqslant 1 \cr
& {\text{Let }}f\left( x \right) = {e^x}{\text{ and }}g\left( x \right) = 0 \cr
& \cr
& {\text{*The mass of the lamina is }} \cr
& m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& m = \rho \int_{ - 1}^1 {\left( {{e^x} - 0} \right)} dx \cr
& m = \rho \int_{ - 1}^1 {{e^x}} dx \cr
& m = \rho \left[ {{e^x}} \right]_{ - 1}^1 \cr
& m = \rho \left( {e - {e^{ - 1}}} \right) \cr
& \cr
& *{\text{The moment about the }}x{\text{ - axis is}} \cr
& {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_x} = \rho \int_{ - 1}^1 {\left[ {\frac{{{e^x}}}{2}} \right]} \left[ {{e^x}} \right]dx \cr
& {M_x} = \frac{\rho }{2}\int_{ - 1}^1 {{e^{2x}}} dx \cr
& {M_x} = \frac{1}{4}\rho \left[ {{e^{2x}}} \right]_{ - 1}^1 \cr
& {M_x} = \frac{1}{4}\rho \left( {{e^2} - {e^{ - 2}}} \right) \cr
& \cr
& *{\text{The moment about the }}y{\text{ - axis is}} \cr
& {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_y} = \rho \underbrace {\int_{ - 1}^1 x \left[ {{e^x}} \right]dx}_{{\text{Integrate by parts}}} \cr
& {M_y} = \rho \underbrace {\int_{ - 1}^1 x {e^x}dx}_{{\text{Integrate by parts}}} \cr
& {M_y} = \rho \left[ {x{e^x} - {e^x}} \right]_{ - 1}^1 \cr
& {M_y} = \rho \left[ {e - e} \right] - \rho \left[ { - {e^{ - 1}} - {e^{ - 1}}} \right] \cr
& {M_y} = 2{e^{ - 1}}\rho \cr
& \cr
& *{\text{The coordinates }}\left( {\overline x ,\overline y } \right){\text{ of the centroid are:}} \cr
& \overline x = \frac{{{M_y}}}{m} = \frac{{2{e^{ - 1}}\rho }}{{\rho \left( {e - {e^{ - 1}}} \right)}} = \frac{{2e}}{{e\left( {{e^2} - 1} \right)}} = \frac{2}{{{e^2} - 1}} \approx 0.313 \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{1}{4}\rho \left( {{e^2} - {e^{ - 2}}} \right)}}{{\rho \left( {e - {e^{ - 1}}} \right)}} = \frac{1}{4}\left( {\frac{{\frac{{{e^4} - 1}}{{{e^2}}}}}{{\frac{{{e^2} - 1}}{e}}}} \right) = \frac{{{e^2} + 1}}{{4e}} \approx 0.77 \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{2}{{{e^2} - 1}},\frac{{{e^2} + 1}}{{4e}}} \right) \cr} $$
