Find the area of the region enclosed by the parabola $x^{2}=y,$ the line $y=x+2$ and the $x-$ axis.

Find the area of the region bounded by the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. (in $\pi$)

Find the area of the circle $4x^{2}+4y^{2}=9$ which is interior to the parabola $x^{2}=4y$.

The area under the curve $y = |\cos x - \sin x|$,$0 \leq x \leq \frac{\pi}{2}$,and above the $x$-axis is

Area of the region bounded by $x^2 = 4y$,the $X$-axis,and the line $x = 3$ is . . . . . . sq. units.

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The area bounded by the curve $y = f(x)$ and the line $y = 3$ is: (in $/3$)