If $\left[t\right]$ denotes the greatest integer $\le t$ , then the value of $\frac{3(\mathrm{e}-1)}{\mathrm{e}}{\int }_1^2{\mathrm{x}}^2{\mathrm{e}}^{\left[\mathrm{x}\right]+\left[{\mathrm{x}}^3\right]}\mathrm{dx}$ is:

${\int }_1^2{\mathrm{x}}^2{\mathrm{e}}^{\left[\mathrm{x}\right]+\left[{\mathrm{x}}^3\right]}\mathrm{dx}={\int }_1^2{\mathrm{x}}^2{\mathrm{e}}^{1+\left[{\mathrm{x}}^3\right]}\mathrm{dx}$
$=\frac{\mathrm{e}}{3}\underset{1}{\overset{8}{\int }}{\mathrm{e}}^{\left[t\right]}\mathrm{dt} \mathrm{where} \mathrm{t}={\mathrm{x}}^3$

$\begin{array}{l}=\frac{\mathrm{e}}{3}\left\{{\int }_1^2 \mathrm{edt}+{\int }_2^3 {\mathrm{e}}^2\mathrm{dt}+\dots \dots ..+{\int }_7^8 {\mathrm{e}}^7\mathrm{dt}\right\}\\ =\frac{\mathrm{e}}{3}\left(\mathrm{e}+{\mathrm{e}}^2+\dots \dots \dots +{\mathrm{e}}^7\right)\end{array}$
$\begin{array}{l}=\frac{\mathrm{e}}{3}\mathrm{e}\left(\frac{{\mathrm{e}}^7-1}{\mathrm{e}-1}\right)\\ \therefore \text{ Given DI=}\frac{3(\mathrm{e}-1)}{\mathrm{e}}.\frac{{\mathrm{e}}^2}{3}\frac{({\mathrm{e}}^7-1)}{(\mathrm{e}-1)}={\mathrm{e}}^8-\mathrm{e}\end{array}$