Sketch the region bounded by the curves $y=x^2\text{and}y=\frac{2}{1+x^2}$. Find the area.

If the area enclosed by curve $y=f\left(x\right)\text{and}y=x^2+2$ between the abscissa $x=2\text{and}x=\alpha ,\alpha >2,is\left({\alpha }^3-4{\alpha }^2+8\right)$ sq. units then find function f(x). It is known that curve y=f(x) lies below the parabola $y=x^2+2.$

The ratio in which the line $x-1=0$ divides the area bounded by the curves $2x+1=\sqrt{4y+1},y=x\text{and}y=2$ is

Find the area enclosed by $y=g\left(x\right),$ x-axis, x=1 and x=37, where g(x) is inverse of $f\left(x\right)=x^3+3x+1$.

Let $O\left(0,0\right),A\left(2,0\right),\text{and}B(1,\frac{1}{\sqrt{3}}$ be the vertices of a triangle. Let R be the region consisting of all theose points P inside $\Delta OAB$ which satisfy $d\left(P,OA\right)\le \text{ min }\left[d\left(P,AB\right)\right],$ where d denotes the distance from the point to the corresponding line. Sketch the region R and find its area.

Find the area of the region bounded by the curves $y=x^2,y=\left|2-x^2\right|,\text{and}y=2,$ which lies to the right of the line x=1.

If the area bounded by $f\left(x\right)=\sqrt{\tan x},y=f\left(c\right),x=0\text{and}x=a,0<c<a<\frac{\pi }{2}$ is minimum then find the value of c.

Find the area bounded by the curve $x^2=y,x^2=-y,\text{and}{y}^2=4x-3.$

Find the area of the region enclosed by the curve $y=\left|x-\frac{1}{x}\right|\left(x>0\right)$ and the line y=2

If $S_0,S_1,S_2,\dots$ are areas bounded by the x-axis and half-wave of the curve $y=\sin \pi \sqrt{x},\text{ then prove that }{S}_0,S_1,S_2,\dots$ are in A.P…