Application of ...

What is the area of the region bounded by the parabola $${ y }^{ 2 }=6(x-1)$$ and $${ y }^{ 2 }=3x$$

Area bounded by the curves $$\displaystyle y = \left[ \frac{x^2}{64} + 2 \right]$$ ([.] denotes the greatest integer function) $$y = x - 1$$ and $$x = 0$$ above the x-axis is

The area bounded by the curves $$x= a \cos^3t, y= a \sin^3 t$$ is 

$$Let\quad f(x)=2-\left| x-1 \right| and\quad g(x)={ \left( x-1 \right)  }^{ 2 },\quad then\quad $$

The area bounded by $$y=\sin ^{ -1 }{ x } , y=\cos ^{ -1 }{ x }$$ and the $$x-axis$$, is given by

The area bounded by the curves $$y={ \left( x-1 \right)  }^{ 2 },y={ \left( x+1 \right)  }^{ 2 }$$ and $$y=\dfrac { 1 }{ 4 }$$ is 

Area common to the circle $$x^{2}+y^{2}=64$$ and the parabola $$y^{2}=4x$$ is

If $$f\left(x\right)=$$max$$\left\{\sin{x},\cos{x},\dfrac{1}{2}\right\}$$, then the area of the region bounded by the curves $$y=f\left(x\right),x-$$axis $$y-$$axis and $$x=2\pi$$ is

The parabola $$y=\dfrac{x^2}{2}$$ divides the circle $$x^2+y^2=8$$ into two parts. Find the area of both parts.

The area of the region lying between the line x-y+2=0 and the curve x=$$\sqrt y $$.