## Calculus (3rd Edition)

$$\left(\frac{7}{6}, 3\right)$$
From the following figure, it is clear that $y=3$ in the center. We calculate the moments as follows: \begin{align*} M_{x}&=\frac{1}{2} \int_{0}^{2}\left((f(x))^{2}-(g(x))^{2}\right) d x\\ &=\frac{1}{2} \int_{0}^{2}\left((x+4)^{2}-(2-x)^{2}\right) d x\\ &=\frac{1}{2}\left(\frac{1}{3}(x+4)^{3}+\frac{1}{3}(2-x)^{3}\right) \bigg|_{0}^{2}\\ &=24 \end{align*} and \begin{align*} M_{y}&=\int_{0}^{2} x(f(x)-g(x)) d x\\ &=\int_{0}^{2} x((x+4)-(2-x)) d x\\ &=\int_{0}^{2} (2x^2+2x) d x\\ &=\frac{2}{3}x^3+x^2\bigg|_{0}^{2}\\ &=\frac{28}{3} \end{align*} and \begin{align*} A&=\int_{0}^{2}((x+4)-(2-x)) d x\\ &=\frac{1}{2}(x+4)^2+\frac{1}{2}(2-x)^2\bigg|_{0}^{2}\\ &=8 \end{align*} Hence $$(\bar{x},\bar{y})= \left(\frac{7}{6}, 3\right)$$