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

解

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

3 - \frac{π}{12} - 1

十進法表記

0.47025 \dots

解答ステップ

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

絶対数を排除する

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

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

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

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

簡素化

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

= 3 - \frac{π}{12} - 1