The value of int_{-1}^{1} x^{2} e^{[x^{3}]} d | vedclass

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(C) Let $I = \int_{-1}^{1} x^{2} e^{[x^{3}]} dx$.We split the integral at $x = 0$:$I = \int_{-1}^{0} x^{2} e^{[x^{3}]} dx + \int_{0}^{1} x^{2} e^{[x^{

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$\frac{e+1}{3e}$

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(C) Let $I = \int_{-1}^{1} x^{2} e^{[x^{3}]} dx$.We split the integral at $x = 0$:$I = \int_{-1}^{0} x^{2} e^{[x^{3}]} dx + \int_{0}^{1} x^{2} e^{[x^{3}]} dx$.For $x \in [-1, 0)$,$x^{3} \in [-1, 0)$,so $[x^{3}] = -1$.For $x \in [0, 1)$,$x^{3} \in [0, 1)$,so $[x^{3}] = 0$.Substituting these values:$I = \int_{-1}^{0} x^{2} e^{-1} dx + \int_{0}^{1} x^{2} e^{0} dx$$I = \frac{1}{e} \int_{-1}^{0} x^{2} dx + \int_{0}^{1} x^{2} dx$$I = \frac{1}{e} \left[ \frac{x^{3}}{3} \right]_{-1}^{0} + \left[ \frac{x^{3}}{3} \right]_{0}^{1}$$I = \frac{1}{e} \left( 0 - \left( -\frac{1}{3} \right) \right) + \left( \frac{1}{3} - 0 \right)$$I = \frac{1}{3e} + \frac{1}{3} = \frac{1+e}{3e}$.

The value of $\int_{-1}^{1} x^{2} e^{[x^{3}]} dx$,where $[t]$ denotes the greatest integer $\leq t$,is

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