integral from 0 to L of x*sin((npix)/L)

解

\int_{0}^{L} x \cdot sin (\frac{nπ x}{L}) d x

- \frac{( - 1 ) ^{n} L ^{2}}{πn}

解答ステップ

\int_{0}^{L} x sin (\frac{nπ x}{L}) d x

部分積分を適用する

= [- L \frac{1}{πn} x cos (\frac{πn x}{L}) - \int - L \frac{1}{πn} cos (\frac{πn x}{L}) d x]_{0}^{L}

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

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

簡素化

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

限度を計算する:

- \frac{( - 1 ) ^{n} L ^{2}}{πn}

= - \frac{( - 1 ) ^{n} L ^{2}}{πn}