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

解答

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

- \frac{5}{4 e ^{2}} + \frac{1}{4}

十进制

0.08083 \dots

求解步骤

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

使用分布积分法

= [- \frac{1}{2} e^{- 2 t} t^{2} - \int - e^{- 2 t} t d t]_{0}^{1}

\int - e^{- 2 t} t d t = - \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t})

= [- \frac{1}{2} e^{- 2 t} t^{2} - (- \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t}))]_{0}^{1}

化简

= [- \frac{1}{2} e^{- 2 t} t^{2} + \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t})]_{0}^{1}

带入上下限计算后得:

- \frac{5}{4 e ^{2}} + \frac{1}{4}

= - \frac{5}{4 e ^{2}} + \frac{1}{4}

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