integral from 0 to 2 of x^2sin((npix)/2)

解答

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

\frac{16 ( - 1 ) ^{n} - 16}{π ^{3} n ^{3}} - \frac{8 ( - 1 ) ^{n}}{πn}

求解步骤

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

使用分布积分法

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

\int - \frac{4}{πn} x cos (\frac{πn}{2} x) d x = - \frac{16}{π ^{3} n ^{3}} (\frac{π}{2} n x sin (\frac{π}{2} n x) + cos (\frac{π}{2} n x))

= [- \frac{2}{πn} x^{2} cos (\frac{πn}{2} x) - (- \frac{16}{π ^{3} n ^{3}} (\frac{π}{2} n x sin (\frac{π}{2} n x) + cos (\frac{π}{2} n x)))]_{0}^{2}

化简

= [- \frac{2}{πn} x^{2} cos (\frac{πn}{2} x) + \frac{16}{π ^{3} n ^{3}} (\frac{π}{2} n x sin (\frac{π}{2} n x) + cos (\frac{π}{2} n x))]_{0}^{2}

带入上下限计算后得:

(- 1)^{n} \frac{16}{π ^{3} n ^{3}} - \frac{8 ( - 1 ) ^{n}}{πn} - \frac{16}{π ^{3} n ^{3}}

= (- 1)^{n} \frac{16}{π ^{3} n ^{3}} - \frac{8 ( - 1 ) ^{n}}{πn} - \frac{16}{π ^{3} n ^{3}}

化简

= \frac{16 ( - 1 ) ^{n} - 16}{π ^{3} n ^{3}} - \frac{8 ( - 1 ) ^{n}}{πn}

\int_{- 1}^{1} (\frac{6}{x ^{2} + 1}) d x

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