Answer
$(\overline {x}, \overline {y})=(\dfrac{6}{5},\dfrac{8}{7} )$
Work Step by Step
We have $\overline {x}=\dfrac{3}{4} \int_{0}^{2} x(2x^2-x^3) dx \\ =\dfrac{3}{4} [ \dfrac{x^4}{2}-\dfrac{x^5}{5}]_{0}^{2} \\= \dfrac{3}{4} \times \dfrac{8}{5}=\dfrac{6}{5}$
Now, $\overline {y}=\dfrac{3}{4} \int_{0}^{2} \dfrac{x^3}{2}(2x^2-x^3) dx \\ =\dfrac{3}{8} [\dfrac{x^6}{3}-\dfrac{x^7}{7}]_{0}^{2} \\=\dfrac{8}{7}$
So, $(\overline {x}, \overline {y})=(\dfrac{6}{5},\dfrac{8}{7} )$
