Area under Simple Curves

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 line $y=x$ partitions the circle ${\left(x-a\right)}^2+y^2=a^2$ in two segments.

What is the area of major segment?

The line $y=x$ partitions the circle ${\left(x-a\right)}^2+y^2=a^2$ in two segments.

What is the area of minor segment?

Find the value of $x_1$ so that ${\int }_0^{x_1}ydx=5$

The area of the region $x,y:x^2\le y\le \left|x^2-4\right|,y\ge 1$ is

The area of the region enclosed by the curve $f\left(x\right)=\max \{\sin x,\cos x\},-\pi \le x\le \pi$ and the $x-$axis is

The area of the region enclosed by the curve $y=x^3$ and its tangent at the point $(–1, –1)$ is

The area of the region $\left\{\left(x,y\right): x^2\le y\le 8-x^2,y\le 7\right\}$ is 

Area bounded by the curve $2y^2=3x$ and the line $x+y=3$ outside the circle ${\left(x-3\right)}^2+y^2=2$ and above $x-$axis is $A$. The value of $4\left(\mathrm{\pi }+4\mathrm{A}\right)$ is

Area of region enclosed by curve $y=x^3$ and its tangent at $\left(-1,-1\right)$.

The area under the curve of equations $: x^2\le y, y\le 8-x^2 \text{and} y\le 7$ is

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 of the region bounded by the curve $y=x^2$ and the line $y=4$.

Shown below is an ellipse whose equation is $\frac{x^2}{81}+\frac{y^2}{36}=1.$

Find the area of the shaded region in terms of $\mathrm{\pi }.$Show your steps.

Find the area enclosed by the circle $x^2+y^2=25$

The area bounded by the curve $y=\left|x^2-9\right|$ and the line $y=3$ is

Find the area(in sq. units) bounded by $y=\left|{\log }_e\left|x\right|\right|$ and $y=0$.

Find the area(in sq. units) bounded by $y={\log }_e\left|x\right|$ and $y=0$.

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

The area of region between the curve $y=4x^2$ and the line $y=6x-2$ is in sq units