Answer
$$\dfrac{3}{2}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Average=\dfrac{1}{\pi} \int \int _R [(1-x)^2+y^2] \space dy \space dx =\dfrac{1}{\pi} \times \int ^{2\pi}_0 \int^{1}_0[(1-rcos\theta)^2+r^2 \sin^2\theta] \space r \space dr \space d\theta $
or, $=\dfrac{1}{\pi} \times \int^{2\pi}_0 \int^1_0 (r^3-2r^2 \cos\theta+r) \times dr \times d\theta $
or, $=\dfrac{1}{\pi} \times \int^{2\pi}_0 (\dfrac{3}{4}-\dfrac{2 \cos\theta}{3}) \space d\theta $
or, $=\dfrac{1}{\pi} \times [\dfrac{3}{4}\theta-\dfrac{2 \sin\theta}{3}]^{2\pi}_0$
or, $=\dfrac{3}{2}$
