integral from 0 to 2 of x*sin((npix)/2)

解答

\int_{0}^{2} x \cdot sin (\frac{nπ x}{2}) d x

- \frac{4 ( - 1 ) ^{n}}{πn}

求解步骤

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

使用换元积分法

= \int_{0}^{πn} \frac{4 u sin ( u )}{π ^{2} n ^{2}} d u

提出常数:

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

= \frac{4}{π ^{2} n ^{2}} \cdot \int_{0}^{πn} u sin (u) d u

使用分布积分法

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) - \int - cos (u) d u]_{0}^{πn}

\int - cos (u) d u = - sin (u)

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) - (- sin (u))]_{0}^{πn}

化简

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) + sin (u)]_{0}^{πn}

带入上下限计算后得:

- π (- 1)^{n} n

= \frac{4}{π ^{2} n ^{2}} (- π (- 1)^{n} n)

化简

= - \frac{4 ( - 1 ) ^{n}}{πn}

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