integral from 0 to 1 of y/(e^{3y)}

解答

\int_{0}^{1} \frac{y}{e ^{3 y}} d y

- \frac{4}{9 e ^{3}} + \frac{1}{9}

十进制

0.08898 \dots

求解步骤

\int_{0}^{1} \frac{y}{e ^{3 y}} d y

使用分布积分法

= [- \frac{1}{3} e^{- 3 y} y - \int - \frac{1}{3} e^{- 3 y} d y]_{0}^{1}

\int - \frac{1}{3} e^{- 3 y} d y = \frac{1}{9} e^{- 3 y}

= [- \frac{1}{3} e^{- 3 y} y - \frac{1}{9} e^{- 3 y}]_{0}^{1}

带入上下限计算后得:

- \frac{4}{9 e ^{3}} + \frac{1}{9}

= - \frac{4}{9 e ^{3}} + \frac{1}{9}

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