Answer
$\left( {\overline x ,\overline y } \right) = \left( {\frac{1}{{\ln 2}},\frac{1}{{4\ln 2}}} \right)$
Work Step by Step
$$\eqalign{
& {\text{From the given graph}}{\text{, we can estimate the centroid at}} \cr
& \left( {1.5,0.3} \right) \cr
& \cr
& y = \frac{1}{x},{\text{ }}1 \leqslant x \leqslant 2 \cr
& {\text{Let }}f\left( x \right) = \frac{1}{x}{\text{ and }}g\left( x \right) = 0 \cr
& \cr
& {\text{*The mass of the lamina is }} \cr
& m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& m = \rho \int_1^2 {\left( {\frac{1}{x}} \right)} dx \cr
& m = \rho \left[ {\ln x} \right]_1^2 \cr
& m = \left( {\ln 2} \right)\rho \cr
& \cr
& *{\text{The moment about the }}x{\text{ - axis is}} \cr
& {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_x} = \rho \int_1^2 {\left[ {\frac{{\frac{1}{x}}}{2}} \right]} \left[ {\frac{1}{x}} \right]dx \cr
& {M_x} = \frac{\rho }{2}\int_1^2 {\frac{1}{{{x^2}}}} dx \cr
& {M_x} = \frac{\rho }{2}\left[ { - \frac{1}{x}} \right]_1^2 \cr
& {M_x} = \frac{\rho }{2}\left[ { - \frac{1}{2} + \frac{1}{1}} \right] \cr
& {M_x} = \frac{\rho }{4} \cr
& \cr
& *{\text{The moment about the }}y{\text{ - axis is}} \cr
& {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_y} = \rho \int_0^2 x \left[ {\frac{1}{x}} \right]dx \cr
& {M_y} = \rho \int_1^2 {dx} \cr
& {M_y} = \rho \left[ x \right]_1^2 \cr
& {M_y} = \rho \left( {2 - 1} \right) \cr
& {M_y} = \rho \cr
& \cr
& *{\text{The coordinates }}\left( {\overline x ,\overline y } \right){\text{ of the centroid are:}} \cr
& \overline x = \frac{{{M_y}}}{m} = \frac{\rho }{{\left( {\ln 2} \right)\rho }} = \frac{1}{{\ln 2}} \approx 1.44 \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{\rho }{4}}}{{\left( {\ln 2} \right)\rho }} = \frac{1}{{4\ln 2}} \approx 0.36 \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{1}{{\ln 2}},\frac{1}{{4\ln 2}}} \right) \cr} $$