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:

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 curve $y^2=4x$, $y$-axis and the line $y=2$ is equal to

The area (in sq. unit) of the region in the first quadrant, enclosed by the $x$-axis, the line $x=\sqrt{3}$ and the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$is equal to

The area of the region enclosed between then curve $y=8x\left(x-1\right)\left(x-2\right)$, and the $x$-axis is equal to

The area of the region bounded by the curve $y^2=4x,$ the $y$-axis and the line $y=3$ is

The area under the curve $y=\frac{e^x}{e^x+e^{-x}}$ between $x=-1$ and $x=1$ is

The area enclosed by the graph of $2\left|x\right|+3\left|y\right|=6$  above the $x$-axis is -

The area bounded by the curve $\left|x\right|+\left|y\right|=1$ 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

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

If $\left(p, 0\right)$, $0<p<\frac{\mathrm{\pi }}{2}$ is a point on the curve $f\left(x\right)=2\sin x\cos x-\sqrt{3}\sin x$ and $A$ is the area bounded by $f\left(x\right)$ and $x$-axis in the first quadrant, then

Area (in sq. unit) bounded by curve $y={\left(x-1\right)}^2\left(x-2\right)$, $x$-axis and the lines $x=1$ and $x=2$ is

Area bounded by the ellipse $2x^2+3y^2=1$ is $\dots \dots \dots$

The area bounded by curve $y=\sin 2x(x=0$ to $x=\pi )$ and $X-$axis is $\dots \dots \dots$

The area (in sq. unit) enclosed by the curve $y=\ln \left(2e+x\right)$ and the coordinate axes is

Let $a, m$ be real numbers such that $0<a<1<m$. Then the area bounded by the curve $y=\frac{1}{x}$ between $x=\frac{a}{m}$ and $x=\frac{1}{m}$ is

Area (in sq. units) bounded by $y=x^2+1, y^2=x$ and the line $x=0$ and $x=1$ is