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受欢迎的微积分 >

integral from 0 to 10 of e^{-2x}

解答

\int_{0}^{10} e^{- 2 x} d x

解答

\frac{e ^{20} - 1}{2 e ^{20}}

+1

十进制

0.49999 \dots

求解步骤

\int_{0}^{10} e^{- 2 x} d x

使用换元积分法

= \int_{0}^{- 20} - \frac{1}{2} e^{u} d u

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

= - \int_{- 20}^{0} - \frac{1}{2} e^{u} d u

提出常数:

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

= - (- \frac{1}{2} \cdot \int_{- 20}^{0} e^{u} d u)

使用积分常见定则:

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

= - (- \frac{1}{2} [e^{u}]_{- 20}^{0})

化简

= \frac{1}{2} [e^{u}]_{- 20}^{0}

带入上下限计算后得:

1 - \frac{1}{e ^{20}}

= \frac{1}{2} (1 - \frac{1}{e ^{20}})

化简

= \frac{e ^{20} - 1}{2 e ^{20}}

作图

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