integral from 1 to 0 of |2x-1|

解

\int_{1}^{0} ∣ 2 x - 1 ∣ d x

- \frac{1}{2}

十進法表記

- 0.5

解答ステップ

\int_{1}^{0} ∣ 2 x - 1 ∣ d x

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

= - \int_{0}^{1} ∣ 2 x - 1 ∣ d x

絶対数を排除する

= - (\int_{0}^{\frac{1}{2}} - 2 x + 1 d x + \int_{\frac{1}{2}}^{1} 2 x - 1 d x)

\int_{0}^{\frac{1}{2}} - 2 x + 1 d x = \frac{1}{4}

\int_{\frac{1}{2}}^{1} 2 x - 1 d x = \frac{1}{4}

= - (\frac{1}{4} + \frac{1}{4})

簡素化

- (\frac{1}{4} + \frac{1}{4}) : - \frac{1}{2}

= - \frac{1}{2}