integral from 0 to N of e^{-st}e^{3t}

解答

\int_{0}^{N} e^{- s t} e^{3 t} d t

\frac{e ^{- N s + 3 N} - 1}{- s + 3}

求解步骤

\int_{0}^{N} e^{- s t} e^{3 t} d t

化简

e^{- s t} e^{3 t} : e^{- s t + 3 t}

= \int_{0}^{N} e^{- s t + 3 t} d t

使用换元积分法

= \int_{0}^{- s N + 3 N} \frac{e ^{u}}{- s + 3} d u

提出常数:

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

= \frac{1}{- s + 3} \cdot \int_{0}^{- s N + 3 N} e^{u} d u

使用积分常见定则:

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

= \frac{1}{- s + 3} [e^{u}]_{0}^{- s N + 3 N}

化简

\frac{1}{- s + 3} [e^{u}]_{0}^{- s N + 3 N} : \frac{[ e ^{u} ] _{0}^{- N s + 3 N}}{- s + 3}

= \frac{[ e ^{u} ] _{0}^{- N s + 3 N}}{- s + 3}

带入上下限计算后得:

e^{- N s + 3 N} - 1

= \frac{e ^{- N s + 3 N} - 1}{- s + 3}

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