Answer
$\overline {x}=\dfrac{3}{2} , \overline {y} =\dfrac{1}{2}$
Work Step by Step
$m_x=\int_{1}^{2} x^2 \cdot x^{-1} \cdot (\dfrac{2}{x^2}) dx=-2[\dfrac{1}{x}]_1^2=1 $ and $m=\int^{1}_{2} x^2 (2x^{-2}) dx=2$
$\implies \overline {y}=\dfrac{m_x}{m}=\dfrac{1}{2} $
$m_y=\int_{1}^{2} x^2 (x) (2 x^{-2}) dx=(x^2)_1^2=3 $ and $m=\int^{1}_{2} x^2 \times (2x^{-2}) dx=2$
$\implies \overline {x}=\dfrac{m_y}{m}=\dfrac{3}{2} $
