If the area lying in the first quadrant and bounded by the circle $x^2+y^2-4x=0$, the parabola $y^2=x$ and the $X$-axis is $A$, then $6A-9\sqrt{3}=$

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 of the region $\{(x, y): |x-1| \leq y \leq \sqrt{5-x^{2}}\}$ is equal to:

The area between the parabolas $x^2 = \frac{y}{4}$ and $x^2 = 9y$ and the straight line $y = 2$ is:

Area enclosed by the curves $y = \ln x$,$y = \ln|x|$,$y = |\ln x|$,and $y = |\ln|x||$ is equal to

The area bounded by the curves $y = (x + 1)^2$,$y = (x - 1)^2$ and the line $y = 0$ is