Answer
$\overline {x} =\dfrac{1}{2}; \overline {y}=4$
Work Step by Step
We have $\overline {x}=\dfrac{6}{125} \int_{-2}^{3} x\cdot (x+6-x^2) dx $
or, $=\dfrac{6}{125} [ \dfrac{x^3}{3}+3x^2-\dfrac{x^4}{4}]_{-2}^4 $
or, $=\dfrac{1}{12}$
and $\overline {y}=\dfrac{6}{125} \int_{-2}^{3} \dfrac{(x+6+x^2)}{2}\cdot (x+6-x^2) dx $
or, $=\dfrac{3}{125} \times [ \dfrac{x^3}{3}+6x^2+36x-\dfrac{x^5}{5}]_{-2}^{3} $
or, $=4$
