integral from 0 to 1 of sin(npix)

解答

\int_{0}^{1} sin (nπ x) d x

\frac{- ( - 1 ) ^{n} + 1}{πn}

求解步骤

\int_{0}^{1} sin (nπ x) d x

使用换元积分法

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

使用积分常见定则:

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

= \frac{1}{πn} [- cos (u)]_{0}^{πn}

带入上下限计算后得:

- (- 1)^{n} + 1

= \frac{1}{πn} (- (- 1)^{n} + 1)

化简

= \frac{- ( - 1 ) ^{n} + 1}{πn}

in v erse l a pl a ce \frac{3}{s} + \frac{24}{s ^{4}}

\frac{\partial}{\partial y} (4 x^{3} y^{2})

in v erse l a pl a ce \frac{1}{( s - 4 ) ^{3}}

\frac{d}{d x} (6 x^{2} - 8 x)

a re a [0, 5], 5 x, \frac{5}{x}, x = 5