integral from 0 to 1 of x(1-x)sin(npix)

解

\int_{0}^{1} x (1 - x) sin (nπ x) d x

\frac{- 2 ( - 1 ) ^{n} + 2}{π ^{3} n ^{3}}

解答ステップ

\int_{0}^{1} x (1 - x) sin (nπ x) d x

部分積分を適用する

= [- \frac{1}{πn} x cos (πn x) (1 - x) - \int - \frac{1}{πn} cos (πn x) (- 2 x + 1) d x]_{0}^{1}

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

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

簡素化

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

限度を計算する:

- \frac{2 ( - 1 ) ^{n}}{π ^{3} n ^{3}} + \frac{2}{π ^{3} n ^{3}}

= - \frac{2 ( - 1 ) ^{n}}{π ^{3} n ^{3}} + \frac{2}{π ^{3} n ^{3}}

簡素化

= \frac{- 2 ( - 1 ) ^{n} + 2}{π ^{3} n ^{3}}