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人気のある微分積分 >

integral from 0 to pi of usin^2(u)

解

\int_{0}^{π} u sin^{2} (u) d u

解

\frac{π ^{2}}{4}

+1

十進法表記

2.46740 \dots

解答ステップ

\int_{0}^{π} u sin^{2} (u) d u

部分積分を適用する

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

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

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

簡素化

[\frac{1}{4} (u (2 u - sin (2 u)) - 4 \cdot \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u)))]_{0}^{π} : [\frac{1}{8} (2 u^{2} - 2 u sin (2 u) - cos (2 u))]_{0}^{π}

= [\frac{1}{8} (2 u^{2} - 2 u sin (2 u) - cos (2 u))]_{0}^{π}

限度を計算する:

\frac{2 π ^{2} - 1}{8} + \frac{1}{8}

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

簡素化

= \frac{π ^{2}}{4}

グラフ

人気の例

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