The area (in sq. units) of the region enclosed between the parabola $y^{2}=2x$ and the line $x+y=4$ is

If the area of the region bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$ is equal to $\alpha$,then $6 \alpha$ equals

Area of the region $\{(x, y): x^2+(y-2)^2 \leq 4, x^2 \geq 2y\}$ is

Find the area bounded on the left by the $y-$axis,below by the $x-$axis,on the right by $x = \frac{\pi}{2}$,above on the left by $y = \cos x$,and above on the right by $y = \sin x$.

Let $A_{1}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and the $y$-axis in the first quadrant. Also,let $A_{2}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$,the $x$-axis and $x = \frac{\pi}{2}$ in the first quadrant. Then ..... .

The area bounded by the curve $y = \begin{cases} x^2, & x < 0 \\ x, & x \geq 0 \end{cases}$ and the line $y = 4$ is