Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{1}{{e - 1}},\frac{{e + 1}}{4}} \right)$$
Work Step by Step
$$\eqalign{
& y = {e^x},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 1,{\text{ }}\rho \left( {x,y} \right) = k \cr
& {\text{The region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant {e^x},{\text{ 0}} \leqslant x \leqslant 1} \right\} \cr
& {\text{The mass }}m{\text{ of the lamina is given by}} \cr
& m = \iint\limits_R {\rho \left( {x,y} \right)dA} \cr
& m = \int_0^1 {\int_0^{{e^x}} k } dydx \cr
& m = k\int_0^1 {\left[ {\int_0^{{e^x}} {dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& m = k\int_0^1 {\left[ y \right]_0^{{e^x}}} dx \cr
& m = k\int_0^1 {{e^x}} dx \cr
& {\text{Integrate}} \cr
& m = k\left[ {{e^x}} \right]_0^1 \cr
& m = k\left( {{e^0} - {e^1}} \right) \cr
& m = k\left( {e - 1} \right) \cr
& \cr
& {\text{*The moment of mass with respect to the }}x{\text{ axis is:}} \cr
& {M_x} = \iint\limits_R {y\rho \left( {x,y} \right)dA},{\text{ then}} \cr
& {M_x} = \int_0^1 {\int_0^{{e^x}} {y\left( k \right)} } dydx \cr
& {M_x} = k\int_0^1 {\left[ {\int_0^{{e^x}} {ydy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_x} = k\int_0^1 {\left[ {\frac{1}{2}{y^2}} \right]_0^{{e^x}}} dx \cr
& {M_x} = k\int_0^1 {\left[ {\frac{1}{2}{{\left( {{e^x}} \right)}^2} - \frac{1}{2}{{\left( 0 \right)}^2}} \right]_0^{{e^x}}} dx \cr
& {M_x} = \frac{k}{2}\int_0^1 {{e^{2x}}} dx \cr
& {\text{Integrate}} \cr
& {M_x} = \frac{k}{2}\left[ {\frac{1}{2}{e^{2x}}} \right]_0^1 \cr
& {M_x} = \frac{k}{4}\left( {{e^2} - {e^0}} \right) \cr
& {M_x} = \frac{k}{4}\left( {{e^2} - 1} \right) \cr
& \cr
& {\text{*The moment of mass with respect to the }}y{\text{ axis is:}} \cr
& {M_y} = \iint\limits_R {x\rho \left( {x,y} \right)dA},{\text{ then}} \cr
& {M_y} = \int_0^1 {\int_0^{{e^x}} {x\left( k \right)} } dydx \cr
& {M_y} = k\int_0^1 {x\left[ {\int_0^{{e^x}} {dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_y} = k\int_0^1 {x\left[ y \right]_0^{{e^x}}} dx \cr
& {M_y} = k\int_0^1 {x{e^x}} dx \cr
& {\text{Integrating by parts we obtain}} \cr
& {M_y} = k\left[ {x{e^x} - {e^x}} \right]_0^1 \cr
& {M_y} = k\left[ {\left( 1 \right){e^1} - {e^1}} \right] - k\left[ {0{e^0} - {e^0}} \right] \cr
& {M_y} = k \cr
& \cr
& {\text{The center of mass is}} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{{{M_y}}}{m},\frac{{{M_x}}}{m}} \right) \cr
& \frac{{{M_y}}}{m} = \frac{k}{{k\left( {e - 1} \right)}} = \frac{1}{{e - 1}} \cr
& \frac{{{M_x}}}{m} = \frac{{\frac{k}{4}\left( {{e^2} - 1} \right)}}{{k\left( {e - 1} \right)}} = \frac{{1\left( {e + 1} \right)\left( {e - 1} \right)}}{{4\left( {e - 1} \right)}} = \frac{{e + 1}}{4} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{1}{{e - 1}},\frac{{e + 1}}{4}} \right) \cr} $$
