Answer
$$\left( {\overline x ,\overline y } \right) \approx \left( {2.177{\text{ }},{\text{ }}0.320} \right)$$
Work Step by Step
$$\eqalign{
& y = x{e^{ - x/2}},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 4 \cr
& {\text{Using the graph shown below}} \cr
& {\text{Let }}f\left( x \right) = x{e^{ - x/2}}{\text{ and }}g\left( x \right) = 0,{\text{ on }}\left[ {0,4} \right] \cr
& \cr
& {\text{*The mass of the lamina is }} \cr
& m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& m = \rho \int_0^4 {\left( {x{e^{ - x/2}} - 0} \right)} dx \cr
& m = \rho \int_0^4 {x{e^{ - x/2}}} dx \cr
& {\text{Using a graphing utility to approximate the integral}} \cr
& m \approx \rho \left( {2.376} \right) \cr
& m \approx 2.376\rho \cr
& \cr
& *{\text{The moment about the }}x{\text{ - axis is}} \cr
& {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_x} = \rho \int_0^4 {\left[ {\frac{{x{e^{ - x/2}}}}{2}} \right]} \left[ {x{e^{ - x/2}}} \right]dx \cr
& {M_x} = \frac{1}{2}\rho \int_0^4 {{x^2}{e^{ - x}}} dx \cr
& {\text{Using a graphing utility to approximate the integral}} \cr
& {M_x} \approx \frac{1}{2}\rho \left( {1.523} \right) \cr
& {M_x} \approx 0.7619\rho \cr
& \cr
& *{\text{The moment about the }}y{\text{ - axis is}} \cr
& {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_y} = \rho \int_0^4 x \left[ {x{e^{ - x/2}}} \right]dx \cr
& {M_y} = \rho \int_0^4 {{x^2}{e^{ - x/2}}} dx \cr
& {M_y} \approx \rho \left( {5.173} \right) \cr
& {M_y} \approx 5.173\rho \cr
& \cr
& *{\text{The center of mass }}\left( {\overline x ,\overline y } \right){\text{ is given by:}} \cr
& \overline x = \frac{{{M_y}}}{m} = \frac{{5.173\rho }}{{2.376\rho }} \approx 2.177 \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{0.7619\rho }}{{2.376\rho }} \approx 0.320 \cr
& \left( {\overline x ,\overline y } \right) \approx \left( {2.177{\text{ }},{\text{ }}0.320} \right) \cr} $$
