Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.3 Center of Mass - Exercises - Page 483: 23

$(\frac{1}{2(e-2)}, \frac{e^{2}-3}{4(e-2)})$

$M_{x}$ = $\frac{1}{2}\int_0^1{[e^{2x}-1}]dx$ = $\frac{e^{2}-3}{4}$ $M_{y}$ = $\int_0^1{x[e^{x}-1}]dx$ = $\frac{1}{2}$ $A$ = $\int_0^1{[e^{x}-1}]dx$ = $e-2$ so the coordinates of the centroid are $(\frac{1}{2(e-2)}, \frac{e^{2}-3}{4(e-2)})$