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

The integral from-1 to 1 of 2xsin(npix) is -(4(-1)^n)/(pin)

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integral from-1 to 1 of 2xsin(npix)

\int_{- 1}^{1} 2 x sin (nπ x) d x

- \frac{4 ( - 1 ) ^{n}}{πn}

Solution steps

\int_{- 1}^{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_{- 1}^{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]_{- 1}^{1}

= 2 [- \frac{1}{πn} x cos (πn x) - \int - \frac{1}{πn} cos (πn x) d x]_{- 1}^{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))]_{- 1}^{1}

Simplify

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

Compute the boundaries:

- (- 1)^{n} \frac{2}{πn}

= 2 (- (- 1)^{n} \frac{2}{πn})

Simplify

= - \frac{4 ( - 1 ) ^{n}}{πn}

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What is the integral from-1 to 1 of 2xsin(npix) ?
The integral from-1 to 1 of 2xsin(npix) is -(4(-1)^n)/(pin)