The area of the region bounded by $y-x=2$ and $x^{2}=y$ is equal to:

If the area enclosed between the curves $y = kx^2$ and $x = ky^2$ $(k > 0)$ is $1$ square unit,then $k$ is

The area bounded by the curve $y = \begin{cases} x^2, & x < 0 \\ x, & x \geq 0 \end{cases}$ and the line $y = 4$ is

The following figure shows the graph of a continuous function $y=f(x)$ on the interval $[1,3]$. The points $A, B, C$ have coordinates $(1,1), (3,2), (2,3)$ respectively,and the lines $l_1$ and $l_2$ are parallel,with $l_1$ being tangent to the curve at $C$. If the area under the graph of $y=f(x)$ from $x=1$ to $x=3$ is $4$ sq units,then the area of the shaded region is

The area of the region bounded by the lines $x=1, x=2$,and the curves $x(y-e^x)=\sin x$ and $2xy=2\sin x+x^3$ is

In the interval $(0, \pi / 2)$,the area lying between the curves $y = \tan x$ and $y = \cot x$ and the $X$-axis is: