## University Calculus: Early Transcendentals (3rd Edition)

$(\overline {x}, \overline {y})=(\dfrac{1}{2},4)$
We have $\overline {x}=\dfrac{6}{125} \int_{-2}{3} x(x+6-x^2) dx \\ =\dfrac{6}{125} [ \dfrac{x^3}{3}+3x^2-\dfrac{x^4}{4}]_{-2}^4 \\= \dfrac{6}{125} \times \dfrac{125}{12}=\dfrac{1}{12}$ Now, $\overline {y}=\dfrac{6}{125} \int_{-2}{3} \dfrac{(x+6+x^2)}{2}(x+6-x^2) dx \\ =\dfrac{3}{125} [ \dfrac{x^3}{3}+6x^2+36x-\dfrac{x^5}{5}]_{-2}^{3} \\= \dfrac{3}{125} \times \dfrac{500}{3}=4$ So, $(\overline {x}, \overline {y})=(\dfrac{1}{2},4)$