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