integral from 0 to 2 of e^{-st}

解答

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

- \frac{e ^{- 2 s} - 1}{s}

求解步骤

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

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

化简

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

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

带入上下限计算后得:

e^{- 2 s} - 1

= - \frac{e ^{- 2 s} - 1}{s}

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