What is the integral from 0 to 1 of x^2sin(nxpi) ?

The integral from 0 to 1 of x^2sin(nxpi) is (2(-1)^n-2)/(pi^3n^3)-((-1)^n)/(pin)

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integral from 0 to 1 of x^2sin(nxpi)

\int_{0}^{1} x^{2} sin (n x π) d x

\frac{2 ( - 1 ) ^{n} - 2}{π ^{3} n ^{3}} - \frac{( - 1 ) ^{n}}{πn}

Solution steps

\int_{0}^{1} x^{2} sin (n x π) d x

Apply Integration By Parts

: [\frac{1}{3} x^{3} sin (πn x) - \int \frac{πn}{3} x^{3} cos (πn x) d x]_{0}^{1}

= [\frac{1}{3} x^{3} sin (πn x) - \int \frac{πn}{3} x^{3} cos (πn x) d x]_{0}^{1}

\int \frac{πn}{3} x^{3} cos (πn x) d x = \frac{1}{3} x^{3} sin (πn x) + \frac{1}{πn} x^{2} cos (πn x) - \frac{2}{π ^{2} n ^{2}} x sin (πn x) - \frac{2}{π ^{3} n ^{3}} cos (πn x)

= [\frac{1}{3} x^{3} sin (πn x) - (\frac{1}{3} x^{3} sin (πn x) + \frac{1}{πn} x^{2} cos (πn x) - \frac{2}{π ^{2} n ^{2}} x sin (πn x) - \frac{2}{π ^{3} n ^{3}} cos (πn x))]_{0}^{1}

Expand

\frac{1}{3} x^{3} sin (πn x) - (\frac{1}{3} x^{3} sin (πn x) + \frac{1}{πn} x^{2} cos (πn x) - \frac{2}{π ^{2} n ^{2}} x sin (πn x) - \frac{2}{π ^{3} n ^{3}} cos (πn x)) : - \frac{1}{πn} x^{2} cos (πn x) + \frac{2}{π ^{2} n ^{2}} x sin (πn x) + \frac{2}{π ^{3} n ^{3}} cos (πn x)

= [- \frac{1}{πn} x^{2} cos (πn x) + \frac{2}{π ^{2} n ^{2}} x sin (πn x) + \frac{2}{π ^{3} n ^{3}} cos (πn x)]_{0}^{1}

Compute the boundaries:

(- 1)^{n} \frac{2}{π ^{3} n ^{3}} - (- 1)^{n} \frac{1}{πn} - \frac{2}{π ^{3} n ^{3}}

= (- 1)^{n} \frac{2}{π ^{3} n ^{3}} - (- 1)^{n} \frac{1}{πn} - \frac{2}{π ^{3} n ^{3}}

Simplify

= \frac{2 ( - 1 ) ^{n} - 2}{π ^{3} n ^{3}} - \frac{( - 1 ) ^{n}}{πn}

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What is the integral from 0 to 1 of x^2sin(nxpi) ?
The integral from 0 to 1 of x^2sin(nxpi) is (2(-1)^n-2)/(pi^3n^3)-((-1)^n)/(pin)