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人気のある微分積分 >

integral from 0 to 2 of 2tsin(2pit)

解

\int_{0}^{2} 2 t sin (2 π t) d t

解

- \frac{2}{π}

+1

十進法表記

- 0.63661 \dots

解答ステップ

\int_{0}^{2} 2 t sin (2 π t) d t

定数を除く:

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

= 2 \cdot \int_{0}^{2} t sin (2 π t) d t

部分積分を適用する

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

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

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

簡素化

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

限度を計算する:

- \frac{1}{π}

= 2 (- \frac{1}{π})

簡素化

= - \frac{2}{π}

グラフ

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