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

The integral from 0 to L of x*sin((npix)/L) is -((-1)^nL^2)/(pin)

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integral from 0 to L of x*sin((npix)/L)

\int_{0}^{L} x \cdot sin (\frac{nπ x}{L}) d x

- \frac{( - 1 ) ^{n} L ^{2}}{πn}

Solution steps

\int_{0}^{L} x sin (\frac{nπ x}{L}) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

- \frac{( - 1 ) ^{n} L ^{2}}{πn}

= - \frac{( - 1 ) ^{n} L ^{2}}{πn}

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What is the integral from 0 to L of x*sin((npix)/L) ?
The integral from 0 to L of x*sin((npix)/L) is -((-1)^nL^2)/(pin)