Answer
$\dfrac{3(1-\sin 1)}{2}$
Work Step by Step
The average value of a function $f(x,y)$ over a region $D$ can be found as: $f_{av}=\dfrac{1}{A} \iint_{D} f(x,y)\ dA$
and $D$ is the projection of the surface on the xy-plane.
The region $D$ using the point of intersection can be expressed as follows:
$$D=\left\{ (x, y) | 0 \leq y \leq x^2 , 0 \leq x \leq 1 \right\}$$
Now, $Area=\iint_{D} f(x,y)\ dA=\int_0^1 \int_0^{x^2} \ dy \ dx$ or, $=[\dfrac{x^3}{3}]_0^1 $ or, $=\dfrac{1}{3}$
Next, $f_{av}= (3) \int_0^1 \int_0^{x^2} x \sin y \ dy \ dx$ or, $=(3) \int_0^1 -x \cos (x^2) +x \ dx $ or, $=\dfrac{3}{2} \int_0^1 [1-\cos (x^2) ] \times 2x \ dx $
Now, set $x^2 =u$ and $2x =du$
So, $f_{av}=\dfrac{3}{2} \int_0^1 (1-u) du \\=\dfrac{3}{2} [u-\sin u ]_0^1 $
or, $=\dfrac{3(1-\sin 1)}{2}$
