integral from 0 to 2 of (t-1)e^{(t-1)^2}

解答

\int_{0}^{2} (t - 1) e^{(t - 1)^{2}} d t

0

求解步骤

\int_{0}^{2} (t - 1) e^{(t - 1)^{2}} d t

使用换元积分法

= \int_{- 1}^{1} e^{u^{2}} u d u

使用分布积分法

= [\frac{1}{2} e^{u^{2}} u^{2} - \int e^{u^{2}} u^{3} d u]_{- 1}^{1}

\int e^{u^{2}} u^{3} d u = \frac{1}{2} (e^{u^{2}} u^{2} - e^{u^{2}})

= [\frac{1}{2} e^{u^{2}} u^{2} - \frac{1}{2} (e^{u^{2}} u^{2} - e^{u^{2}})]_{- 1}^{1}

化简

[\frac{1}{2} e^{u^{2}} u^{2} - \frac{1}{2} (e^{u^{2}} u^{2} - e^{u^{2}})]_{- 1}^{1} : [\frac{1}{2} e^{u^{2}}]_{- 1}^{1}

= [\frac{1}{2} e^{u^{2}}]_{- 1}^{1}

带入上下限计算后得:

0

= 0

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