integral from 0 to 3 of xsin((mpix)/3)

解

\int_{0}^{3} x sin (\frac{mπ x}{3}) d x

\frac{9 sin ( πm ) - 9 πm cos ( πm )}{π ^{2} m ^{2}}

解答ステップ

\int_{0}^{3} x sin (\frac{mπ x}{3}) d x

u置換積分法を適用する

= \int_{0}^{m \cdot 3 π} \frac{u sin ( \frac{u}{3} )}{π ^{2} m ^{2}} d u

定数を除く:

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

= \frac{1}{π ^{2} m ^{2}} \cdot \int_{0}^{m \cdot 3 π} u sin (\frac{u}{3}) d u

部分積分を適用する

= \frac{1}{π ^{2} m ^{2}} [- 3 u cos (\frac{u}{3}) - \int - 3 cos (\frac{u}{3}) d u]_{0}^{3 πm}

\int - 3 cos (\frac{u}{3}) d u = - 9 sin (\frac{u}{3})

= \frac{1}{π ^{2} m ^{2}} [- 3 u cos (\frac{u}{3}) - (- 9 sin (\frac{u}{3}))]_{0}^{3 πm}

簡素化

= \frac{1}{π ^{2} m ^{2}} [- 3 u cos (\frac{u}{3}) + 9 sin (\frac{u}{3})]_{0}^{3 πm}

限度を計算する:

9 sin (πm) - 9 πm cos (πm)

= \frac{1}{π ^{2} m ^{2}} (9 sin (πm) - 9 πm cos (πm))

簡素化

= \frac{9 sin ( πm ) - 9 πm cos ( πm )}{π ^{2} m ^{2}}