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受欢迎的微积分 >

integral from 0 to 5 of 0.05te^{-0.35t}

解答

\int_{0}^{5} 0.05 t e^{- 0.35 t} d t

解答

0.21311 \dots

求解步骤

\int_{0}^{5} 0.05 t e^{- 0.35 t} d t

提出常数:

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

= 0.05 \cdot \int_{0}^{5} t e^{- 0.35 t} d t

使用换元积分法

= 0.05 \cdot \int_{0}^{- 1.75} \frac{20 e ^{u} u}{2.45} d u

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

= 0.05 (- \int_{- 1.75}^{0} \frac{20 e ^{u} u}{2.45} d u)

提出常数:

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

= 0.05 (- \frac{20}{2.45} \cdot \int_{- 1.75}^{0} e^{u} u d u)

化简

= 0.05 (- 8.16326 \dots \cdot \int_{- 1.75}^{0} e^{u} u d u)

使用分布积分法

= 0.05 (- 8.16326 \dots [e^{u} u - \int e^{u} d u]_{- 1.75}^{0})

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

= 0.05 (- 8.16326 \dots [e^{u} u - e^{u}]_{- 1.75}^{0})

化简

= - 0.40816 \dots [e^{u} u - e^{u}]_{- 1.75}^{0}

带入上下限计算后得:

- 0.52212 \dots

= - 0.40816 \dots (- 0.52212 \dots)

化简

= 0.21311 \dots

作图

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