integral from 0 to x of λe^{-λx}

解答

\int_{0}^{x} λ e^{- λ x} d x

- e^{- λ x} + 1

求解步骤

\int_{0}^{x} λ e^{- λ x} d x

提出常数:

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

= λ \cdot \int_{0}^{x} e^{- λ x} d x

使用换元积分法

= λ \cdot \int_{0}^{- λ x} - \frac{e ^{u}}{λ} d u

提出常数:

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

= λ (- \frac{1}{λ} \cdot \int_{0}^{- λ x} e^{u} d u)

使用积分常见定则:

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

= λ (- \frac{1}{λ} [e^{u}]_{0}^{- λ x})

化简

λ (- \frac{1}{λ} [e^{u}]_{0}^{- λ x}) : - [e^{u}]_{0}^{- λ x}

= - [e^{u}]_{0}^{- λ x}

带入上下限计算后得:

e^{- λ x} - 1

= - (e^{- λ x} - 1)

化简

= - e^{- λ x} + 1

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