$$I = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $$

$$ = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{1 + [{x^3}]}}dx} $$ ($$\because$$ $$[x] = 1$$ when $$x \in (12)$$)

$$ = 3(e - 1)\int_1^2 {{x^2}{e^{[{x^3}]}}dx} $$

Let $${x^3} = t$$

$$I = (e - 1)\int_1^8 {{e^{[t]}}dt} $$

$$ = ({e^{ - 1}})({e^1} + {e^2} + {e^3}\, + \,...\, + \,{e^7})$$

$$ = (e - 1)e{{({e^7} - 1)} \over {e - 1}}$$

$$ = {e^8} - e$$