integral from 0 to pi of x^2sin(kx)

解

\int_{0}^{π} x^{2} sin (k x) d x

\frac{- π ^{2} k ^{2} cos ( πk ) + 2 ( πk sin ( πk ) + cos ( πk ) ) - 2}{k ^{3}}

解答ステップ

\int_{0}^{π} x^{2} sin (k x) d x

部分積分を適用する

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

\int - \frac{2}{k} x cos (k x) d x = - \frac{2}{k ^{3}} (k x sin (k x) + cos (k x))

= [\frac{1}{k} (- x^{2} cos (k x) - k (- \frac{2}{k ^{3}} (k x sin (k x) + cos (k x))))]_{0}^{π}

簡素化

[\frac{1}{k} (- x^{2} cos (k x) - k (- \frac{2}{k ^{3}} (k x sin (k x) + cos (k x))))]_{0}^{π} : [\frac{1}{k} (- x^{2} cos (k x) + \frac{2}{k ^{2}} (k x sin (k x) + cos (k x)))]_{0}^{π}

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

限度を計算する:

\frac{1}{k} (- π^{2} cos (kπ) + \frac{2}{k ^{2}} (kπ sin (kπ) + cos (kπ))) - \frac{2}{k ^{3}}

= \frac{1}{k} (- π^{2} cos (kπ) + \frac{2}{k ^{2}} (kπ sin (kπ) + cos (kπ))) - \frac{2}{k ^{3}}

簡素化

= \frac{- π ^{2} k ^{2} cos ( πk ) + 2 ( πk sin ( πk ) + cos ( πk ) ) - 2}{k ^{3}}