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人気のある微分積分 >

integral from 0 to 1 of 9cos(2x)

解

\int_{0}^{1} 9 cos (2 x) d x

解

\frac{9}{2} sin (2)

+1

十進法表記

4.09183 \dots

解答ステップ

\int_{0}^{1} 9 cos (2 x) d x

定数を除く:

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

= 9 \cdot \int_{0}^{1} cos (2 x) d x

u置換積分法を適用する

= 9 \cdot \int_{0}^{2} cos (u) \frac{1}{2} d u

定数を除く:

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

= 9 \cdot \frac{1}{2} \cdot \int_{0}^{2} cos (u) d u

共通積分を使用する:

\int cos (u) d u = sin (u)

= 9 \cdot \frac{1}{2} [sin (u)]_{0}^{2}

簡素化

9 \cdot \frac{1}{2} [sin (u)]_{0}^{2} : \frac{9}{2} [sin (u)]_{0}^{2}

= \frac{9}{2} [sin (u)]_{0}^{2}

限度を計算する:

sin (2)

= \frac{9}{2} sin (2)

グラフ

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