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人気のある微分積分 >

integral from 0 to pi of 5x^2sin(1/2 x)

解

\int_{0}^{π} 5 x^{2} sin (\frac{1}{2} x) d x

解

5 (8 π - 16)

+1

十進法表記

45.66370 \dots

解答ステップ

\int_{0}^{π} 5 x^{2} sin (\frac{1}{2} x) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int_{0}^{π} x^{2} sin (\frac{1}{2} x) d x

簡素化

\frac{1}{2} x : \frac{x}{2}

= 5 \cdot \int_{0}^{π} x^{2} sin (\frac{x}{2}) d x

部分積分を適用する

= 5 [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

\int - 4 x cos (\frac{x}{2}) d x = - 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))

= 5 [- 2 x^{2} cos (\frac{x}{2}) - (- 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2})))]_{0}^{π}

簡素化

= 5 [- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))]_{0}^{π}

限度を計算する:

8 π - 16

= 5 (8 π - 16)

グラフ

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