integral from 0 to b of xe^{-ax}

解答

\int_{0}^{b} x e^{- a x} d x

\frac{- ab e ^{- ab} - e ^{- ab} + 1}{a ^{2}}

求解步骤

\int_{0}^{b} x e^{- a x} d x

使用换元积分法

= \int_{0}^{- ab} \frac{e ^{u} u}{a ^{2}} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{a ^{2}} \cdot \int_{0}^{- ab} e^{u} u d u

使用分布积分法

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

\int e^{u} d u = e^{u}

= \frac{1}{a ^{2}} [e^{u} u - e^{u}]_{0}^{- ab}

化简

\frac{1}{a ^{2}} [e^{u} u - e^{u}]_{0}^{- ab} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- ab}}{a ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- ab}}{a ^{2}}

带入上下限计算后得:

- ab e^{- ab} - e^{- ab} + 1

= \frac{- ab e ^{- ab} - e ^{- ab} + 1}{a ^{2}}

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