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