Area Bounded by Curves

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)

Let $n\ge 2$ be a natural number and $f:\left[0, 1\right]\to \mathrm{ℝ}$ be the function defined by

$f\left(x\right)=\left\{\begin{array}{ccc}n\left(1-2nx\right)& \mathrm{if}& 0\le x\le \frac{1}{2n}\\ 2n\left(2nx-1\right)& \mathrm{if}& \frac{1}{2n}\le x\le \frac{3}{4n}\\ 4n\left(1-nx\right)& \mathrm{if}& \frac{3}{4n}\le x\le \frac{1}{n}\\ \frac{n}{n-1}\left(nx-1\right)& \mathrm{if}& \frac{1}{n}\le x\le 1\end{array}\right.$

If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f\left(x\right)$ is $4$, then the maximum value of the function $f$ is

Let $f:\left[0, 1\right]\to \left[0, 1\right]$ be the function defined by $f\left(x\right)=\frac{x^3}{3}-x^2+\frac{5}{9}x+\frac{17}{36}$. Consider the square region $S=\left[0, 1\right]\times \left[0, 1\right]$. Let $G=\left\{\left(x, y\right)\in S:y>f\left(x\right)\right\}$ be called the green region and $R=\left\{\left(x, y\right)\in S:y<f\left(x\right)\right\}$ be called the red region. Let $L_h=\left\{\left(x, h\right)\in S:x\in \left[0, 1\right]\right\}$ be the horizontal line drawn at a height $h\in \left[0, 1\right]$. Then which of the following statements is(are) true?

Draw a rough sketch of the graph $y=\left|x+2\right|+\left|x-2\right|$ for $-3\le x\le 3$ and hence find the enclosed area.

Find the area bounded by $R=\left\{\left(x,y\right):\max \left\{0,\ln x\right\}\le y\le 2^x,\frac{1}{2}\le x\le 2\right\}$.

A curve $C$ is given by ${\left[f\left(y\right)\right]}^{\frac{2}{3}}+{\left[f\left(x\right)\right]}^{\frac{1}{3}}=0,$ satisfy the equation $\left(x-y\right)f\left(x+y\right)+\left(x+y\right)f\left(x-y\right)=4xy\left(x^2-y^2\right)$, Then the area of the region bounded by curve $C$ and the line $x=-3$, in sq.units is $A$. Find $\frac{A^2}{2}$

Consider the curve $x^2=4\left[\sqrt{x}\right]y$, where $\left[·\right]$ denote the greatest integer function. Then the area of the region enclosed by the given curve and the $x-$axis from ordinates $x=1,x=9$ is:

Find the area bounded by $y={\sin }^{-1}\left(\sin x\right)$ and $x$-axis for $x\in \left[0,100\pi \right]$.

Find the volume of solid obtained by revolving the curve $x^2=y-3$ about $x-\mathrm{axis}$ from $x=1$ to $x=3$.

The area bounded by a curve, the axis of co-ordinates and the ordinate of some point of the curve is equal to the length of the corresponding arc of the curve. If the curve passes through the point $P\left(0,1\right)$, then the equation of this curve can be

If the area bounded by the curves $y=x^2$ and $y=\frac{2}{1+x^2}$ is $\lambda$ sq. units and the area bounded by the region $\left[x\right]\left[y\right]=2$ is $\mu$ sq. units, then the correct relation is (where $\left[·\right]$ denotes greatest integer function)

Let $a,b,c$ are real and distinct numbers and $f(x)$ is a quadratic function such that $\left[\begin{array}{ccc}4a^2& 4a& 1\\ 4b^2& 4b& 1\\ 4c^2& 4c& 1\end{array}\right]\left[\begin{array}{c}f\left(-1\right)\\ f\left(1\right)\\ f\left(2\right)\end{array}\right]=\left[\begin{array}{l}3a^2+3a\\ 3b^2+3b\\ 3c^2+3c\end{array}\right]$. Point $A$ and $B$ are such that $A$ is the point where $y=f\left(x\right)$ cuts the $x$-axis and $B$ is a point such that $AB$ subtends a right angle at $V$ (local maxima of $f\left(x\right)$. Find the area between the curve $y=f\left(x\right)$ and the chord $AB$.