The area of the region (in sq. units) bounded by the curves $x^2+y^2=16$ and $y^2=6x$ is

The area enclosed by $y=\sqrt{5-x^{2}}$ and $y=|x-1|$ is

If the area of the bounded region $R=\{(x, y): \max \{0, \log _{e} x\} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2\}$ is $\alpha(\log _{e} 2)^{-1}+\beta(\log _{e} 2)+\gamma$,then the value of $(\alpha+\beta-2 \gamma)^{2}$ is equal to:

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 bounded by $y=x+1$, $y=\cos x$ and the $X$-axis is

The area bounded by the curves $y=2x^2$, $y=\max \{x-[x], x+|x|\}$ and the lines $x=0, x=2$ (in sq units) is