The area in the first quadrant between $x^2 + y^2 = \pi^2$ and $y = \sin x$ is

The area of the figure bounded by the parabolas $y^2+8x=16$ and $y^2-24x=48$ is

If the area of the region bounded by $y=\cos x$, $y=\sin x$, $x=\frac{\pi}{4}$ and $x=\pi$ is bisected by the line $x=a$, then $\sin \left(a+\frac{\pi}{4}\right)=$

$A$ line passing through the point $A(-2, 0)$ touches the parabola $P: y^2 = x - 2$ at the point $B$ in the first quadrant. The area of the region bounded by the line $AB$,the parabola $P$,and the $x$-axis is:

The area bounded by the curves $y^2 = 4x$ and $x^2 = 4y$ is

Given: $f(x) = \begin{cases} x, & 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, & x = \frac{1}{2} \\ 1-x, & \frac{1}{2} < x \leq 1 \end{cases}$ and $g(x) = (x-\frac{1}{2})^2, x \in R$. Then the area (in sq. units) of the region bounded by the curves $y=f(x)$ and $y=g(x)$ between the lines $2x=1$ and $2x=\sqrt{3}$ is: