If the area enclosed between the curves $y^2 = 4kx$ and $y = kx$ for $k > 0$ is $\frac{2}{3}$ sq. units, then $k =$ ?

The area (in sq. units) of the region $\{ (x,y) : x \ge 0, x + y \le 3, x^2 \le 4y \text{ and } y \le 1 + \sqrt{x} \}$ is:

The area of the region bounded by the parabola $(y-2)^{2}=(x-1)$,the tangent to it at the point whose ordinate is $3$,and the $x$-axis is:

If the area of the region $\{(x, y): x^{2/3} + y^{2/3} \leq 1, x + y \geq 0, y \geq 0\}$ is $A$,then find the value of $\frac{256A}{\pi}$.

The ratio in which the $x$-axis divides the area of the region bounded by the curves $y = x^2 - 4x$ and $y = 2x - x^2$ is:

If $A$ is the area bounded by the curves $y = |\cos x|$ and $y = 5 - \frac{4}{\pi} |x - \pi|$ in $x \in [0, 2\pi]$,then the value of $\left( \frac{A}{2} + 2 \right)$ is