If the area of the region bounded by the curves $y^2-2y=-x$ and $x+y=0$ is $A$,then $8A$ is equal to

The area of the region between the curves $y=\sqrt{\frac{1+\sin x}{\cos x}}$ and $y=\sqrt{\frac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\frac{\pi}{4}$ is

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

The following figure shows the graph of a continuous function $y=f(x)$ on the interval $[1,3]$. The points $A, B, C$ have coordinates $(1,1), (3,2), (2,3)$ respectively,and the lines $l_1$ and $l_2$ are parallel,with $l_1$ being tangent to the curve at $C$. If the area under the graph of $y=f(x)$ from $x=1$ to $x=3$ is $4$ sq units,then the area of the shaded region is

Area of the region (in square units) enclosed by the curves $y^2=8(x+2)$,$y^2=4(1-x)$ and the $Y$-axis is

If $f(x)$ is a continuous,increasing,and odd function such that $\int_{-1}^{4} f(x) \,dx = 10$ and $\int_{0}^{1} f(x) \,dx = \frac{3}{2}$,then the area bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = -4$ and $x = 4$ is: