integral from 0 to 1 of 2^{-θ}

解答

\int_{0}^{1} 2^{- θ} d θ

\frac{1}{2 ln ( 2 )}

十进制

0.72134 \dots

求解步骤

\int_{0}^{1} 2^{- θ} d θ

使用换元积分法

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

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

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

提出常数:

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

= - (- \int_{- 1}^{0} 2^{u} d u)

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

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

化简

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

带入上下限计算后得:

\frac{1}{2 ln ( 2 )}

= \frac{1}{2 ln ( 2 )}

x \to 0 lim (3 x - 4)

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\frac{d y}{d t} = \frac{t}{t ^{2} y + y}

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 6)^{4})

\int x^{\frac{- 1}{2}} d x