integral from 0 to 1 of 5^{-θ}

解答

\int_{0}^{1} 5^{- θ} d θ

\frac{4}{5 ln ( 5 )}

十进制

0.49706 \dots

求解步骤

\int_{0}^{1} 5^{- θ} d θ

使用换元积分法

= \int_{0}^{- 1} - 5^{u} d u

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

= - \int_{- 1}^{0} - 5^{u} d u

提出常数:

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

= - (- \int_{- 1}^{0} 5^{u} d u)

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - (- [\frac{5 ^{u}}{ln ( 5 )}]_{- 1}^{0})

化简

= [\frac{5 ^{u}}{ln ( 5 )}]_{- 1}^{0}

带入上下限计算后得:

\frac{4}{5 ln ( 5 )}

= \frac{4}{5 ln ( 5 )}

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