Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.3 Center of Mass - Exercises - Page 483: 22

$(\frac{2}{6\ln{2}-3}, \frac{1}{12\ln{2}-6})$

$M_{x}$ = $\frac{1}{2}\int_1^2{[(\frac{1}{x})^{2}-(2-x)^{2}]}dx$ = $\frac{1}{12}$ $M_{y}$ = $\int_1^2{x[(\frac{1}{x})-(2-x)]}dx$ = $\frac{1}{3}$ $A$ = $\int_1^2{[(\frac{1}{x})-(2-x)]}dx$ = $\ln{2}-\frac{1}{2}$ so the coordinates of the centroid are $(\frac{2}{6\ln{2}-3}, \frac{1}{12\ln{2}-6})$