integral from 0 to 2 of ye^{-xy}

解答

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

- e^{- 2 y} + 1

求解步骤

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

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

化简

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

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

带入上下限计算后得:

e^{- 2 y} - 1

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

化简

= - e^{- 2 y} + 1

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