$A_1$ represents the area between two lines $L_r\equiv (m+r)!x$ and $L_{r+1}\equiv (m+(r+1\left)\right)!x$ and line $x = 2$. $(m\ge 0)$

$\frac{A_0}{A_1}=\frac{\frac{1}{2}\times 2\left[\right(m+1)!-m!]}{\frac{1}{2}\times 2\times \left[\right(m+2)!-(m+1)!]}$

$\Rightarrow \frac{A_0}{A_1}=\frac{m}{m^2+2m+1}=\frac{1}{m+\frac{1}{m}+2}$

$\mathrm{For}{\left(\frac{A_0}{A_1}\right)}_{max}=\frac{1}{2+2}=\frac{1}{4}$

$A_0=\frac{1}{2}\times 2xm!\left(m\right)$

$A_1=\frac{1}{2}\times 2x(m+1)!(m+1)$

$A_2=\frac{1}{2}\times 2x(m+2)!(m+2)$

$A_0<A_1<A_2.......$

$y_0=m!x$

For minimum area $m = 0.1$

$A=\frac{1}{2}\times 2\times 2=2$