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

解答

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

- e^{- t} + 1

求解步骤

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

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

带入上下限计算后得:

e^{- t} - 1

= - (e^{- t} - 1)

化简

= - e^{- t} + 1

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