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

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

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

\int_{0}^{1} x \cdot sin (nπ x) d x

- \frac{( - 1 ) ^{n}}{πn}

Solution steps

\int_{0}^{1} x sin (nπ x) d x

Apply Integration By Parts

: [- \frac{1}{πn} x cos (πn x) - \int - \frac{1}{πn} cos (πn x) d x]_{0}^{1}

= [- \frac{1}{πn} x cos (πn x) - \int - \frac{1}{πn} cos (πn x) d x]_{0}^{1}

\int - \frac{1}{πn} cos (πn x) d x = - \frac{1}{π ^{2} n ^{2}} sin (πn x)

= [- \frac{1}{πn} x cos (πn x) - (- \frac{1}{π ^{2} n ^{2}} sin (πn x))]_{0}^{1}

Simplify

= [- \frac{1}{πn} x cos (πn x) + \frac{1}{π ^{2} n ^{2}} sin (πn x)]_{0}^{1}

Compute the boundaries:

- (- 1)^{n} \frac{1}{πn}

= - (- 1)^{n} \frac{1}{πn}

Simplify

= - \frac{( - 1 ) ^{n}}{πn}

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\int_{0}^{1} e^{2 x + 3} d x

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