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

解

\int_{0}^{π} sin (x) - \frac{1}{2} d x

23 - \frac{π}{6} - 2

十進法表記

0.94050 \dots

解答ステップ

\int_{0}^{π} sin (x) - \frac{1}{2} d x

絶対数を排除する

= \int_{0}^{\frac{π}{6}} - sin (x) + \frac{1}{2} d x + \int_{\frac{π}{6}}^{\frac{5 π}{6}} sin (x) - \frac{1}{2} d x + \int_{\frac{5 π}{6}}^{π} - sin (x) + \frac{1}{2} d x

\int_{0}^{\frac{π}{6}} - sin (x) + \frac{1}{2} d x = \frac{π}{12} + \frac{3}{2} - 1

\int_{\frac{π}{6}}^{\frac{5 π}{6}} sin (x) - \frac{1}{2} d x = 3 - \frac{π}{3}

\int_{\frac{5 π}{6}}^{π} - sin (x) + \frac{1}{2} d x = - 1 + \frac{3}{2} + \frac{π}{12}

= \frac{π}{12} + \frac{3}{2} - 1 + 3 - \frac{π}{3} - 1 + \frac{3}{2} + \frac{π}{12}

簡素化

\frac{π}{12} + \frac{3}{2} - 1 + 3 - \frac{π}{3} - 1 + \frac{3}{2} + \frac{π}{12} : 23 - \frac{π}{6} - 2

= 23 - \frac{π}{6} - 2