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人気のある微分積分 >

integral from 0 to 1/2 of 5xcos(pix)

解

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

解

5 (\frac{1}{2 π} - \frac{1}{π ^{2}})

+1

十進法表記

0.28916 \dots

解答ステップ

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

部分積分を適用する

= 5 [\frac{1}{π} (x sin (π x) - π \cdot \int \frac{1}{π} sin (π x) d x)]_{0}^{\frac{1}{2}}

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

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

簡素化

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

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

限度を計算する:

\frac{1}{2 π} - \frac{1}{π ^{2}}

= 5 (\frac{1}{2 π} - \frac{1}{π ^{2}})

グラフ

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