integral from 0 to pi/4 of xsin(2x)

解

\int_{0}^{\frac{π}{4}} x sin (2 x) d x

\frac{1}{4}

十進法表記

0.25

解答ステップ

\int_{0}^{\frac{π}{4}} x sin (2 x) d x

部分積分を適用する

= [\frac{1}{2} (- x cos (2 x) - 2 \cdot \int - \frac{1}{2} cos (2 x) d x)]_{0}^{\frac{π}{4}}

\int - \frac{1}{2} cos (2 x) d x = - \frac{1}{4} sin (2 x)

= [\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{\frac{π}{4}}

簡素化

[\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{\frac{π}{4}} : [\frac{1}{2} (- x cos (2 x) + \frac{1}{2} sin (2 x))]_{0}^{\frac{π}{4}}

= [\frac{1}{2} (- x cos (2 x) + \frac{1}{2} sin (2 x))]_{0}^{\frac{π}{4}}

限度を計算する:

\frac{1}{4}

= \frac{1}{4}