integral from 0 to pi of 2xsin(kx)

解

\int_{0}^{π} 2 x sin (k x) d x

\frac{2 ( sin ( πk ) - πk cos ( πk ) )}{k ^{2}}

解答ステップ

\int_{0}^{π} 2 x sin (k x) d x

定数を除く:

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

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

部分積分を適用する

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

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

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

簡素化

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

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

限度を計算する:

\frac{1}{k} (\frac{1}{k} sin (πk) - π cos (πk))

= 2 \cdot \frac{1}{k} (\frac{1}{k} sin (πk) - π cos (πk))

簡素化

= \frac{2 ( sin ( πk ) - πk cos ( πk ) )}{k ^{2}}