integral from 0 to pi/2 of-xcos(x)+x

解

\int_{0}^{\frac{π}{2}} - x cos (x) + x d x

- \frac{π}{2} + \frac{π ^{2}}{8} + 1

十進法表記

0.66290 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} - x cos (x) + x d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{0}^{\frac{π}{2}} x d x = \frac{π ^{2}}{8}

= - (\frac{π}{2} - 1) + \frac{π ^{2}}{8}

簡素化

= - \frac{π}{2} + \frac{π ^{2}}{8} + 1