The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to: (in $/3$)

The area (in square units) of the region enclosed between the parabola $y^2=2x$ and the line $y=4x-1$ is:

Let the area of the region enclosed by the curves $y=3x$,$2y=27-3x$,and $y=3x-x\sqrt{x}$ be $A$. Then $10A$ is equal to

The area of the region $S = \{(x, y) : y^{2} \leq 8x, y \geq \sqrt{2}x, x \geq 1\}$ is

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 between the curves $y=\sqrt{\frac{1+\sin x}{\cos x}}$ and $y=\sqrt{\frac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\frac{\pi}{4}$ is