Question 205 of 246MediumMCQ
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 :
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Official solution
\[I = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} \]
\[ = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{1 + [{x^3}]}}dx} \] (\[\because\] \[[x] = 1\] when \[x \in (12)\])
\[ = 3(e - 1)\int_1^2 {{x^2}{e^{[{x^3}]}}dx} \]
Let x3=t
\[I = (e - 1)\int_1^8 {{e^{[t]}}dt} \]
\[ = ({e^{ - 1}})({e^1} + {e^2} + {e^3}\, + \,...\, + \,{e^7})\]
\[ = (e - 1)e{{({e^7} - 1)} \over {e - 1}}\]
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