Area under the Curve

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

The area bounded by the parabolas  $y={(x+1)}^2$  and  $y={(x-1)}^2$  and the line y = 1/4 is:

If area bounded by the curves  $y=f\left(x\right)$ (which lies above $x$-axis $\forall x\ge 1$), the $x$-axis and the ordinates  $x=1 \& x=b$ is $\left[\left(b-1\right)\sin \left(3b+4\right)\right] \text{sq. units} \left(\forall b>1\right)$, then $f\left(x\right)$ is

Let the area bounded by the $x$-axis, curve $y=\left(1+\frac{8}{x^2}\right)$ and the ordinates $x=2$ and $x=4$ is "$A$" sq. unit and if the ordinate $x=\mathrm{a}$ divides the area into two equal parts, then the correct statement among the following is

The area of the region bounded by the curves $y=\sqrt{5-x^2}$ and $y=\left|x-1\right|$ is (in sq. units)

The area bounded by the curve $y=x^2+1$ and the tangents to it drawn from the origin, is

The area bounded by the curves $y=\frac{1}{4}\left|4-x^2\right|$ and $y=7-\left|x\right|$ 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, $\frac{\mathrm{area}\left(PQRS\right)}{\mathrm{area}\left(T\right)}$ equals

Let $f:\left[-1,1\right]\to R$ be a continuous function. Consider the region $S=\left\{\left(x,y\right):-1\le x\le 1 \mathrm{and} 0\le y\le f\left(x\right)\right\}$.

For which one of the following functions $f,$ the area of the region $S$ is the largest?

If the line $x=a$ bisects the area under the curve $y=\frac{1}{x^2}, 1\le x\le 9,$ then a is equal to

If the line, $y=mx$, bisects the area $\text{(in sq. units)}$ of the region $\left\{\left(x,y\right):0\le x\le \frac{3}{2},0\le y\le 1+4x-x^2\right\}$, then $m$ equals

The area (in sq. units) of the region, bounded by the curve $y=\sqrt{x}$ and the lines $y=0, y=x-2$, is

The area bounded by the curves $y^2=12x$ and $x^2=12y$ is divided by the line $x=3$ in two parts. The area (in sq units) of the larger part is

The area (in sq. units) of the region bounded by the curve $\sqrt{x}+\sqrt{y}=1, x, y\ge 0$, and the tangent to it at the point $\left(\frac{1}{4},\frac{1}{4}\right)$ is

The area (in sq. units) of the region bounded by the curve $12y=36-x^2$ and the tangents drawn to it at the points, where the curve intersects the $X$-axis is

The area of region bounded by the lines $y=mx,x=1$ and $x=2$ and the $x$-axis is $7.5$ sq. units, then $m$ is