integral from 0 to 1 of (x^2-x)sin(npix)

解

\int_{0}^{1} (x^{2} - 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^{2} - x) sin (nπ x) d x

拡張

(x^{2} - x) sin (nπ x) : x^{2} sin (πn x) - x sin (πn x)

= \int_{0}^{1} x^{2} sin (πn x) - x sin (πn x) d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} x^{2} sin (πn x) d x - \int_{0}^{1} x sin (πn x) d x

\int_{0}^{1} x^{2} sin (πn x) d x = \frac{2 ( - 1 ) ^{n} - 2}{π ^{3} n ^{3}} - \frac{( - 1 ) ^{n}}{πn}

\int_{0}^{1} x sin (πn x) d x = - \frac{( - 1 ) ^{n}}{πn}

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

簡素化

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