The area of the region $S = \{(x, y): 3x^{2} \leq 4y \leq 6x + 24\}$ is $......$

Area of the region (in sq units) bounded by the curves $y=\sqrt{x}$, $x=\sqrt{y}$ and the lines $x=1$, $x=4$ 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 area (in sq. units) of the region $\{(x, y): 0 \leq y \leq x^{2}+1, 0 \leq y \leq x+1, \frac{1}{2} \leq x \leq 2\}$ is

Let $AB$ be the latus rectum of the parabola $y^2 = 4ax$ in the $XY$-plane. Let $T$ be the region bounded by the finite arc $AB$ of the parabola and the line segment $AB$. $A$ rectangle $PQRS$ of maximum possible area is inscribed in $T$ with $P, Q$ on line $AB$,and $R, S$ on arc $AB$. Then,$\frac{\text{area}(PQRS)}{\text{area}(T)}$ equals

If $f(x)$ is a continuous,increasing,and odd function such that $\int_{-1}^{4} f(x) \,dx = 10$ and $\int_{0}^{1} f(x) \,dx = \frac{3}{2}$,then the area bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = -4$ and $x = 4$ is: