integral from 0 to 1 of-xsin(npix)

解答

\int_{0}^{1} - x sin (nπ x) d x

\frac{( - 1 ) ^{n}}{πn}

求解步骤

\int_{0}^{1} - x sin (nπ x) d x

提出常数:

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

= - \int_{0}^{1} x sin (nπ x) d x

使用分布积分法

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

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

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

化简

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

带入上下限计算后得:

- (- 1)^{n} \frac{1}{πn}

= - (- (- 1)^{n} \frac{1}{πn})

化简

= \frac{( - 1 ) ^{n}}{πn}

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