The area (in sq. units) of the region bounded by the parabola $y = x^2 + 2$ and the lines $y = x + 1$,$x = 0$,and $x = 3$ is:

The area (in sq. units) bounded by the curves $y=\frac{8}{x}$,$y=2x$ and $x=4$ is

The parabolas $y^2 = 4x$ and $x^2 = 4y$ divide the square region bounded by the lines $x = 4$,$y = 4$ and the coordinate axes. If $S_1, S_2, S_3$ are respectively the areas of these parts numbered from top to bottom,then $S_1:S_2:S_3$ 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

The area bounded by the curves $y = \cos x$ and $y = \sin x$ between the ordinates $x = 0$ and $x = \frac{3\pi}{2}$ 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