Let $y=g\left(x\right)$ be the inverse of a bijective mapping $f:R\to R$, where $f\left(x\right)=3x^3+2x$. The area bounded by graph of $g\left(x\right),$ the $x$-axis and the ordinate at $x=5$ 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 

$R_1=\left\{\left(x, y\right) : 0\le x\le 1 \& 0\le y\le 1\right\}$ and $R_2=\left\{\left(x, y\right) : x^2+y^2\le 4/3\right\}$.

The area of the region $R_1\cap R_2$ can be expressed as $\frac{a\sqrt{3}+b\pi }{9}$, where $a$ and $b$ are integers, then -