The area included between the parabola $y=\frac{x^2}{4 a}$ and the curve $y=\frac{8 a^3}{x^2+4 a^2}$ is

Find the area bounded by the curves $x = \sqrt{y - 1}$ and $y = x + 1$.

The area of the region bounded by $x = 0, y = 0, x = 2, y = 2, y \leq e^x$ and $y \geq \log x$ is

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{\left(1+x^2\right)}}$ with $y(0)=0$. Then the area enclosed by the curve $f(x)=y(x) e^{-\frac{1}{\left(1+x^2\right)}}$ and the line $y=x/4+2$ is:

Find the area of the region enclosed between the two circles $:$ $x^{2}+y^{2}=4$ and $(x-2)^{2}+y^{2}=4$.

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