integral from 0 to l of sin((npix)/l)

解答

\int_{0}^{l} sin (\frac{nπ x}{l}) d x

\frac{l ( - ( - 1 ) ^{n} + 1 )}{πn}

求解步骤

\int_{0}^{l} sin (\frac{nπ x}{l}) d x

使用换元积分法

= \int_{0}^{nπ l} \frac{sin ( \frac{u}{l} )}{πn} d u

提出常数:

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

= \frac{1}{πn} \cdot \int_{0}^{nπ l} sin (\frac{u}{l}) d u

使用换元积分法

= \frac{1}{πn} \cdot \int_{0}^{πn} sin (v) l d v

提出常数:

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

= \frac{1}{πn} l \cdot \int_{0}^{πn} sin (v) d v

使用积分常见定则:

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

= \frac{1}{πn} l [- cos (v)]_{0}^{πn}

带入上下限计算后得:

- (- 1)^{n} + 1

= \frac{1}{πn} l (- (- 1)^{n} + 1)

化简

= \frac{l ( - ( - 1 ) ^{n} + 1 )}{πn}

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