Let the area of the region bounded by the curve $y=\max\{\sin x, \cos x\}$, lines $x=0, x=\frac{3\pi}{2}$ and the x-axis be $A$. Then, $A+A^{2}$ is equal to:

The area of the region (in sq. units) enclosed between the curves $y=|x|$,$y=[x]$ and the ordinates $x=-1$,$x=0$,$x=1$ is

The area (in square units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1, x=0$ and $x=3$,is

One of the points of intersection of the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(\ell \sqrt{5}+m)-n \log_{e}(1+\sqrt{5})$,where $\ell, m, n \in N$. Then $\ell+m+n$ is equal to

The area (in $sq. \, units$) of the region bounded by the curves $x^{2}+2y-1=0$,$y^{2}+4x-4=0$,and $y^{2}-4x-4=0$ in the upper half plane is $....$

The area (in sq. units) bounded by the curves $x=y^2$ and $x=3-2y^2$ is