integral from 0 to pi/2 of x^2sin(2x)

解

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

\frac{π ^{2} - 2}{8} - \frac{1}{4}

十進法表記

0.73370 \dots

解答ステップ

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

部分積分を適用する

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

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

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

簡素化

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

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

限度を計算する:

\frac{π ^{2} - 2}{8} - \frac{1}{4}

= \frac{π ^{2} - 2}{8} - \frac{1}{4}