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

解答

\int_{0}^{t} e^{- 3 u} d u

- \frac{1}{3} (e^{- 3 t} - 1)

求解步骤

\int_{0}^{t} e^{- 3 u} d u

使用换元积分法

= \int_{0}^{- 3 t} - \frac{1}{3} e^{v} d v

提出常数:

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

= - \frac{1}{3} \cdot \int_{0}^{- 3 t} e^{v} d v

使用积分常见定则:

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

= - \frac{1}{3} [e^{v}]_{0}^{- 3 t}

带入上下限计算后得:

e^{- 3 t} - 1

= - \frac{1}{3} (e^{- 3 t} - 1)

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