The area of the region for which $0<y<3-2x-x^2$ and $x>0$ is - 

If $C_1\equiv y=\frac{1}{1+x^2}$ and $C_2\equiv y=\frac{x^2}{2}$ be two curves lying in $xy$ plane. Then 

$y=f\left(x\right)$  is a function which satisfies

$\left(\mathrm{i}\right) f\left(0\right)=0$ 

$\left(\mathrm{ii}\right) f^{\text{'}\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)$ and

$\left(\mathrm{iii}\right) f^{\text{'}}\left(0\right)=1$.

Then, the area bounded by the graph of  $y=f\left(x\right),$ the lines $x=0,$ $x-1=0$ and $y+1=0,$ is

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