"If [t] denotes the greatest integer $\le \mathrm{t}$, then the value of ${{3(e - 1)} \over e}\int\limits_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx}$ is :"

Question

Accepted Answer

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

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

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

Let ${x^3} = t$

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

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

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

$= {e^8} - e$

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If [t] denotes the greatest integer $\le \mathrm{t}$, then the value of ${{3(e - 1)} \over e}\int\limits_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx}$ is :