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受欢迎的微积分 >

integral from 0 to 10 of e^{-0.04t}

解答

\int_{0}^{10} e^{- 0.04 t} d t

解答

8.24199 \dots

求解步骤

\int_{0}^{10} e^{- 0.04 t} d t

使用换元积分法

= \int_{0}^{- 0.4} - \frac{1}{0.04} e^{u} d u

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

= - \int_{- 0.4}^{0} - \frac{1}{0.04} e^{u} d u

提出常数:

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

= - (- \frac{1}{0.04} \cdot \int_{- 0.4}^{0} e^{u} d u)

化简

= - (- 25 \cdot \int_{- 0.4}^{0} e^{u} d u)

使用积分常见定则:

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

= - (- 25 [e^{u}]_{- 0.4}^{0})

化简

= 25 [e^{u}]_{- 0.4}^{0}

带入上下限计算后得:

0.32967 \dots

= 25 \cdot 0.32967 \dots

化简

= 8.24199 \dots

作图

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