Let  $f\left(x\right)=\left\{\begin{array}{l}0 if x\le 0\\ P\left(x\right) if 0<x<1\\ 1 if x\ge 1\end{array}\right.$

where$P\left(x\right)$ is a polynomial of least degree such that $f\left(x\right)$ is a differentiable function for all $x\in R$. 

Area bounded by the graph of $y=f\left(x\right),x=0, and x=1$ 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.
 $\therefore P^{/}\left(x\right)=kx\left(x-1\right)\Rightarrow P\left(x\right)=k\left(\frac{x^3}{3}-\frac{x^2}{2}\right)+C$
Since,$P\left(0\right)=0\Rightarrow C=0$ and $P\left(1\right)=1\Rightarrow k=-6$ 
Hence, $P\left(x\right)=-2x^3+3x^2$.
$f\left(x\right)=\left\{\begin{array}{l}0 if x\le 0\\ -2x^3+3x^2 if 0<x<1\\ 1 if x\ge 1\end{array}\right.$