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

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

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

\int_{0}^{1} (x - 1) sin (nπ x) d x

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

Solution steps

\int_{0}^{1} (x - 1) sin (nπ x) d x

Expand

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

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

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

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

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

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