What is the integral from 0 to 1 of 2xsin(npix) ?

The integral from 0 to 1 of 2xsin(npix) is -(2(-1)^n)/(pin)

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integral from 0 to 1 of 2xsin(npix)

\int_{0}^{1} 2 x sin (nπ x) d x

- \frac{2 ( - 1 ) ^{n}}{πn}

Solution steps

\int_{0}^{1} 2 x sin (nπ x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \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}

= 2 [- \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)

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

Simplify

= 2 [- \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}

= 2 (- (- 1)^{n} \frac{1}{πn})

Simplify

= - \frac{2 ( - 1 ) ^{n}}{πn}

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What is the integral from 0 to 1 of 2xsin(npix) ?
The integral from 0 to 1 of 2xsin(npix) is -(2(-1)^n)/(pin)