integral from-1 to 1 of-xsin(npix)

解答

\int_{- 1}^{1} - x sin (nπ x) d x

\frac{2 ( - 1 ) ^{n}}{πn}

求解步骤

\int_{- 1}^{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_{- 1}^{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]_{- 1}^{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))]_{- 1}^{1}

化简

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

带入上下限计算后得:

- (- 1)^{n} \frac{2}{πn}

= - (- (- 1)^{n} \frac{2}{πn})

化简

= \frac{2 ( - 1 ) ^{n}}{πn}

\int_{2}^{5} \frac{3}{x} d x

\int_{0}^{1} (e^{2 x} - 5) d x

\int_{0}^{y} 2 e^{- 2 x} d x

\int_{1}^{\infty} \frac{1}{2 x + 6} d x

\int_{1}^{2} \frac{x ^{6} + 1}{2 x ^{3}} d x