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受欢迎的微积分 >

integral from 0 to 1 of 6e^{-6sqrt(x)}

解答

\int_{0}^{1} 6 e^{- 6 x} d x

解答

- \frac{- e ^{6} + 7}{3 e ^{6}}

+1

十进制

0.32754 \dots

求解步骤

\int_{0}^{1} 6 e^{- 6 x} d x

提出常数:

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

= 6 \cdot \int_{0}^{1} e^{- 6 x} d x

使用换元积分法

= 6 \cdot \int_{0}^{1} e^{- 6 u} \cdot 2 u d u

提出常数:

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

= 6 \cdot 2 \cdot \int_{0}^{1} e^{- 6 u} u d u

使用换元积分法

= 6 \cdot 2 \cdot \int_{0}^{- 6} \frac{e ^{v} v}{36} d v

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

= 6 \cdot 2 (- \int_{- 6}^{0} \frac{e ^{v} v}{36} d v)

提出常数:

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

= 6 \cdot 2 (- \frac{1}{36} \cdot \int_{- 6}^{0} e^{v} v d v)

使用分布积分法

= 6 \cdot 2 (- \frac{1}{36} [e^{v} v - \int e^{v} d v]_{- 6}^{0})

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

= 6 \cdot 2 (- \frac{1}{36} [e^{v} v - e^{v}]_{- 6}^{0})

化简

6 \cdot 2 (- \frac{1}{36} [e^{v} v - e^{v}]_{- 6}^{0}) : - \frac{1}{3} [e^{v} v - e^{v}]_{- 6}^{0}

= - \frac{1}{3} [e^{v} v - e^{v}]_{- 6}^{0}

带入上下限计算后得:

- 1 + \frac{7}{e ^{6}}

= - \frac{1}{3} (- 1 + \frac{7}{e ^{6}})

化简

= - \frac{- e ^{6} + 7}{3 e ^{6}}

作图

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