Find the area lying between the curves $y^{2}=2x$ and $y=x$.

The area of the region bounded by the curve $y = \max \{| x |, x | x - 2 |\}$,the $x$-axis,and the lines $x = -2$ and $x = 4$ is equal to . . . . . . .

The area bounded by the curve $y=\cos x$,the line joining $(-\pi / 4, \cos (-\pi / 4))$ and $(0,2)$ and the line joining $(\pi / 4, \cos (\pi / 4))$ and $(0,2)$ 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

Area of the region bounded by two parabolas $y=x^{2}$ and $x=y^{2}$ is

Let $C_{1}$ be the curve obtained by the solution of the differential equation $2xy \frac{dy}{dx} = y^{2} - x^{2}, x > 0$. Let the curve $C_{2}$ be the solution of $\frac{2xy}{x^{2} - y^{2}} = \frac{dy}{dx}$. If both the curves pass through $(1, 1)$,then the area enclosed by the curves $C_{1}$ and $C_{2}$ is equal to: