What is the integral from 0 to 2 of xsin((npi)/2 x) ?

The integral from 0 to 2 of xsin((npi)/2 x) is -(4(-1)^n)/(pin)

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integral from 0 to 2 of xsin((npi)/2 x)

\int_{0}^{2} x sin (\frac{nπ}{2} x) d x

- \frac{4 ( - 1 ) ^{n}}{πn}

Solution steps

\int_{0}^{2} x sin (\frac{nπ}{2} x) d x

\frac{nπ}{2} = nπ \frac{1}{2}

= \int_{0}^{2} x sin (nπ \frac{1}{2} x) d x

Apply u-substitution

: \int_{0}^{πn} \frac{4 u sin ( u )}{π ^{2} n ^{2}} d u

= \int_{0}^{πn} \frac{4 u sin ( u )}{π ^{2} n ^{2}} d u

Take the constant out:

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

= \frac{4}{π ^{2} n ^{2}} \cdot \int_{0}^{πn} u sin (u) d u

Apply Integration By Parts

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

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) - \int - cos (u) d u]_{0}^{πn}

\int - cos (u) d u = - sin (u)

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) - (- sin (u))]_{0}^{πn}

Simplify

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) + sin (u)]_{0}^{πn}

Compute the boundaries:

- π (- 1)^{n} n

= \frac{4}{π ^{2} n ^{2}} (- π (- 1)^{n} n)

Simplify

= - \frac{4 ( - 1 ) ^{n}}{πn}

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What is the integral from 0 to 2 of xsin((npi)/2 x) ?
The integral from 0 to 2 of xsin((npi)/2 x) is -(4(-1)^n)/(pin)