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

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

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

\int_{- 1}^{0} (x + 1) sin (nπ x) d x

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

Solution steps

\int_{- 1}^{0} (x + 1) sin (nπ x) d x

Expand

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

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

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

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

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

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