What is the integral from 0 to 2 of xsin(npix) ?

The integral from 0 to 2 of xsin(npix) is (sin(2pin))/(pi^2n^2)-(2cos(2pin))/(pin)

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integral from 0 to 2 of xsin(npix)

\int_{0}^{2} x sin (nπ x) d x

\frac{sin ( 2 πn )}{π ^{2} n ^{2}} - \frac{2 cos ( 2 πn )}{πn}

Solution steps

\int_{0}^{2} 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}^{2}

\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}^{2}

Simplify

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

Compute the boundaries:

\frac{1}{π ^{2} n ^{2}} sin (2 πn) - \frac{2}{πn} cos (2 πn)

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

Simplify

= \frac{sin ( 2 πn )}{π ^{2} n ^{2}} - \frac{2 cos ( 2 πn )}{πn}

What is the integral from 0 to 2 of xsin(npix) ?
The integral from 0 to 2 of xsin(npix) is (sin(2pin))/(pi^2n^2)-(2cos(2pin))/(pin)