Let $f\left(x\right)=\left\{\begin{array}{ccc}0& if& x\le 0\\ P\left(x\right)& if& 0<x<1\\ 1& if& x\ge 1\end{array}\right\}$  where P(x) is a polynomial of least degree such that f(x) is a differentiable function for all $x\in R$ . Area bounded by the graph of y = f(x), with x–axis and between the ordinates  x = 0 and x = 1 is
 

Since, f(x) is differentiable and hence continuous  $\forall x\in R$
$\Rightarrow f\left(0^{+}\right)=f\left(0^{-}\right)\Rightarrow P\left(0\right)=0$  and
 $\Rightarrow f\left(0^{+}\right)=f^{\mathbf{/}}\left(0^{-}\right)\Rightarrow P\left(0\right)=0$
Similarly, continuity at $x=1\Rightarrow P\left(1\right)=1$  and differentiability at  $x=1\Rightarrow P\left(1\right)=0$.
Since, P(x) is a polynomial of least degree and $P^{/}\left(x\right)$  vanishes x = 1 and x = 0.
Hence, P(x) must be cubic.
 $$\begin{gathered} \therefore P^{/}\left(x\right)=kx(x-1) \\ \Rightarrow P\left(x\right)=k(\frac{x^3}{3}-\frac{x^2}{2})+C \end{gathered}$$
 Since,  $P\left(0\right)=0\Rightarrow C=0$
and  $P\left(1\right)=1\Rightarrow k=-6$
Hence,  $P\left(x\right)=-2x^3+3x^2$.
 $\therefore f\left(x\right)=\left\{\begin{array}{ccc}0& if& x\le 0\\ -2x^3+3x^2& if& 0<x<1\\ 1& if& x\ge 1\end{array}\right\}$