integral from 0 to 5 of e^{-st}(2t)

解答

\int_{0}^{5} e^{- s t} (2 t) d t

\frac{2 ( - 5 s e ^{- 5 s} - e ^{- 5 s} + 1 )}{s ^{2}}

求解步骤

\int_{0}^{5} e^{- s t} \cdot 2 t d t

提出常数:

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

= 2 \cdot \int_{0}^{5} e^{- s t} t d t

使用换元积分法

= 2 \cdot \int_{0}^{- s \cdot 5} \frac{e ^{u} u}{s ^{2}} d u

提出常数:

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

= 2 \cdot \frac{1}{s ^{2}} \cdot \int_{0}^{- s \cdot 5} e^{u} u d u

使用分布积分法

= 2 \cdot \frac{1}{s ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- 5 s}

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

= 2 \cdot \frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- 5 s}

化简

2 \cdot \frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- 5 s} : \frac{2 [ e ^{u} u - e ^{u} ] _{0}^{- 5 s}}{s ^{2}}

= \frac{2 [ e ^{u} u - e ^{u} ] _{0}^{- 5 s}}{s ^{2}}

带入上下限计算后得:

- 5 s e^{- 5 s} - e^{- 5 s} + 1

= \frac{2 ( - 5 s e ^{- 5 s} - e ^{- 5 s} + 1 )}{s ^{2}}

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