integral from 0 to 3 of te^{-0.6t}

解答

\int_{0}^{3} t e^{- 0.6 t} d t

1.49211 \dots

求解步骤

\int_{0}^{3} t e^{- 0.6 t} d t

使用换元积分法

= \int_{0}^{- 1.8} \frac{5 e ^{u} u}{1.8} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1.8}^{0} \frac{5 e ^{u} u}{1.8} d u

提出常数:

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

= - \frac{5}{1.8} \cdot \int_{- 1.8}^{0} e^{u} u d u

化简

= - 2.77777 \dots \cdot \int_{- 1.8}^{0} e^{u} u d u

使用分布积分法

= - 2.77777 \dots [e^{u} u - \int e^{u} d u]_{- 1.8}^{0}

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

= - 2.77777 \dots [e^{u} u - e^{u}]_{- 1.8}^{0}

带入上下限计算后得:

- 0.53716 \dots

= - 2.77777 \dots (- 0.53716 \dots)

化简

= 1.49211 \dots

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