What is the integral from 0 to 1 of (1-x)sin(npix) ?

The integral from 0 to 1 of (1-x)sin(npix) is (-(-1)^n+1)/(pin)+((-1)^n)/(pin)

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integral from 0 to 1 of (1-x)sin(npix)

\int_{0}^{1} (1 - x) sin (nπ x) d x

\frac{- ( - 1 ) ^{n} + 1}{πn} + \frac{( - 1 ) ^{n}}{πn}

Solution steps

\int_{0}^{1} (1 - x) sin (nπ x) d x

Expand

(1 - x) sin (nπ x) : sin (πn x) - x sin (πn x)

= \int_{0}^{1} sin (πn x) - x sin (πn x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} sin (πn x) d x - \int_{0}^{1} x sin (πn x) d x

\int_{0}^{1} sin (πn x) d x = \frac{- ( - 1 ) ^{n} + 1}{πn}

\int_{0}^{1} x sin (πn x) d x = - \frac{( - 1 ) ^{n}}{πn}

= \frac{- ( - 1 ) ^{n} + 1}{πn} - (- \frac{( - 1 ) ^{n}}{πn})

Simplify

= \frac{- ( - 1 ) ^{n} + 1}{πn} + \frac{( - 1 ) ^{n}}{πn}

\int_{\frac{π}{2}}^{\frac{3 π}{2}} x sin (x) d x

\int_{2}^{3} \frac{x}{1 - x ^{2}} d x

\int_{2}^{6} x d x

\int_{1}^{3} (2 x - 1) d x

\int_{- 1}^{3} 2 x - x^{2} d x

What is the integral from 0 to 1 of (1-x)sin(npix) ?
The integral from 0 to 1 of (1-x)sin(npix) is (-(-1)^n+1)/(pin)+((-1)^n)/(pin)