Application of ...

The area (in sq units) of the region $$\{ (x,y):{ y }^{ 2 }\ge 2x$$ and $${ x }^{ 2 }+{ y }^{ 2 }\le 4x ,\chi \ge 0, Y\ge 0\}$$ 

The area bounded by the curve $$xy^{2}=1$$ and the lines $$x=1$$, $$x=2$$ is

The area in square units bounded by the curves $$y = x ^ { 3 } , y = x ^ { 2 }$$ and the ordinates $$x = 1 , x = 2$$ is

The area bounded by the curves $$y=\sqrt{-x}$$ and $$x=-\sqrt{-y}$$, where $$x,y\le0$$, is equal to

Tangents are drawn from a point $$P$$ to a parabola $$y^{2}=4ax$$. The area enclosed by the tangents and the corresponding chord of contact is $$4a^{2}$$. Then point $$P$$ satisfies

Let $$T$$ be the triangle with vertices $$\left (0,0\right), \left (0,{c}^{2}\right)\ and \left (c,{c}^{2}\right)$$ and let $$R$$ be the region between $$y=cx$$ and $$y={x}^{2}\ where c>0$$ then 

Let $$f(x)=minimum (x+1,\sqrt{1-x}) $$ for all $$x \le 1$$. Then the area bounded by $$y=f(x)$$ and the x-axis is

The area of the figure bounded by the curves $$y ^ { 2 } = 2 x + 1$$ and $$x - y - 1 = 0$$ is 

Area (in $$sq.$$ unit) of region bounded by $$y=2\cos x,\ y=3\tan x$$ and $$y-$$axis is

Let $$f\left( x \right)$$ be a non-negative continuous function such that the area bounded by the curve $$y= f\left( x \right) $$, x-axis and the ordinates $$x=\cfrac { \pi  }{ 4 } $$, $$x=\beta >\cfrac { \pi  }{ 4 } $$ is $$\left( \beta \sin { \beta  } +\cfrac { \pi  }{ 4 } \cos { \beta  } +\sqrt { 2 } \beta -\cfrac { \pi  }{ 2 }  \right) $$. Then $$f\left( \cfrac { \pi  }{ 2 }  \right) $$ is