Answer
$$\dfrac{1}{2 \pi} $$
Work Step by Step
$I_{avg}=(\dfrac{1}{\frac{\pi}{4}}) \times \int_{0}^{1} \int_{0}^{\sqrt {1-x^2}} x \times y \ dy \ dx \\=\int_{0}^{1} \dfrac{4}{\pi} \times [\dfrac{xy^2}{2}]_{0}^{\sqrt {1-x^2}} \ dx \\= \int_{0}^{1} \dfrac{4}{\pi} \times [\dfrac{xy^2}{2}]_{0}^{\sqrt {1-x^2}} \ dx \\=\int_{0}^{1}\dfrac{2(x-x^3)}{\pi} \ dx \\= [\dfrac{2}{\pi} \dfrac{x^2}{2}]_{0}^{1} - [\dfrac{2}{\pi} \dfrac{x^3}{3}]_{0}^{1} \ dx \\=\dfrac{1}{2 \pi} $$
