integral from 0 to x of 2xe^{-x^2}

解答

\int_{0}^{x} 2 x e^{- x^{2}} d x

- e^{- x^{2}} + 1

求解步骤

\int_{0}^{x} 2 x e^{- x^{2}} d x

提出常数:

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

= 2 \cdot \int_{0}^{x} x e^{- x^{2}} d x

使用换元积分法

= 2 \cdot \int_{0}^{- x^{2}} - \frac{e ^{u}}{2} d u

提出常数:

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

= 2 (- \frac{1}{2} \cdot \int_{0}^{- x^{2}} e^{u} d u)

使用积分常见定则:

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

= 2 (- \frac{1}{2} [e^{u}]_{0}^{- x^{2}})

化简

2 (- \frac{1}{2} [e^{u}]_{0}^{- x^{2}}) : - [e^{u}]_{0}^{- x^{2}}

= - [e^{u}]_{0}^{- x^{2}}

带入上下限计算后得:

e^{- x^{2}} - 1

= - (e^{- x^{2}} - 1)

化简

= - e^{- x^{2}} + 1

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