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

解

\int_{0}^{1} ∣ x (2 x - 1) ∣ d x

\frac{1}{4}

十進法表記

0.25

解答ステップ

\int_{0}^{1} ∣ x (2 x - 1) ∣ d x

絶対数を排除する

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

\int_{0}^{\frac{1}{2}} - 2 x^{2} + x d x = \frac{1}{24}

\int_{\frac{1}{2}}^{1} x (2 x - 1) d x = \frac{5}{24}

= \frac{1}{24} + \frac{5}{24}

簡素化

\frac{1}{24} + \frac{5}{24} : \frac{1}{4}

= \frac{1}{4}