If I = ∫ 1 2 d x 2 x 3 - 9 x 2 + 12 x + 4 , then

$f\left(x\right)=\frac{1}{\sqrt{2x^3-9x^2+12x+4}}$
$f^{\text{'}}\left(x\right)=\frac{-1}{2}\frac{\left(6x^2-18x+12\right)}{{\left(2x^3-9x^2+12x+4\right)}^{\frac{3}{2}}}$

$f^{\text{'}}\left(x\right)=-\frac{6\left(x-1\right)\left(x-2\right)}{2{\left(2x^3-9x^2+12x+4\right)}^{\frac{3}{2}}}$

$f\text{'}\left(x\right)=0 \Rightarrow x=1, 2$
$f\left(1\right)=\frac{1}{3}$ , $\mathrm{}f\left(2\right)=\frac{1}{\sqrt{8}}$

$\frac{1}{3}<f\left(x\right)<\frac{1}{\sqrt{8}}$

$\Rightarrow {\int }_1^2\frac{1}{3}dx<{\int }_1^{2}f\left(x\right)dx<{\int }_1^2\frac{1}{\sqrt{8}}dx$
$\Rightarrow \frac{1}{3}<I<\frac{1}{\sqrt{8}}$

$\Rightarrow \frac{1}{9}<I^2<\frac{1}{8}$.