integral from 0 to t of e^{-ax}

解答

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

- \frac{e ^{- a t} - 1}{a}

求解步骤

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

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

化简

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

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

带入上下限计算后得:

e^{- a t} - 1

= - \frac{e ^{- a t} - 1}{a}

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