If [ t ] denotes the greatest integer t , then the value of 3 ( e - 1 ) e 1 2 x 2 e [ x ] + [ x 3 ] ? d x is [2023]

Question

If $[t]$ denotes the greatest integer $\leq t$ , then the value of $\frac{3 (e - 1)}{e} \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$ is [2023]

Smart Achievers · Accepted Answer

(4)

Let $I = \frac{3 (e - 1)}{e} \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$

Between 1 and 2, $[x]$ = 1, so $e^{[x]} = e$ .

$∴$ $I = 3 (e - 1) \int_{1}^{2} x^{2} \cdot e^{[x^{3}]} d x$

Let $x^{3} = t \Rightarrow x^{2} d x = \frac{d t}{3}$

$I = \frac{3 (e - 1)}{3} \int_{1}^{8} e^{[t]} d t = (e - 1) \int_{1}^{8} e^{[t]} d t$

$= (e - 1) [e + e^{2} + e^{3} + e^{4} + e^{5} + e^{6} + e^{7}]$ $= (e - 1) [\frac{e (e^{7} - 1)}{e - 1}]$

$\Rightarrow e (e^{7} - 1) = e^{8} - e$

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