integral from 0 to 2 of x^2cos(npix)

解答

\int_{0}^{2} x^{2} cos (nπ x) d x

\frac{4 π ^{2} n ^{2} sin ( 2 πn ) - 2 ( sin ( 2 πn ) - 2 πn cos ( 2 πn ) )}{π ^{3} n ^{3}}

求解步骤

\int_{0}^{2} x^{2} cos (nπ x) d x

使用换元积分法

= \int_{0}^{nπ 2} \frac{u ^{2} cos ( u )}{π ^{3} n ^{3}} d u

提出常数:

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

= \frac{1}{π ^{3} n ^{3}} \cdot \int_{0}^{nπ 2} u^{2} cos (u) d u

使用分布积分法

= \frac{1}{π ^{3} n ^{3}} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{nπ 2}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

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

带入上下限计算后得:

4 π^{2} n^{2} sin (2 πn) - 2 (sin (2 πn) - 2 πn cos (2 πn))

= \frac{1}{π ^{3} n ^{3}} (4 π^{2} n^{2} sin (2 πn) - 2 (sin (2 πn) - 2 πn cos (2 πn)))

化简

= \frac{4 π ^{2} n ^{2} sin ( 2 πn ) - 2 ( sin ( 2 πn ) - 2 πn cos ( 2 πn ) )}{π ^{3} n ^{3}}

\int_{- \infty}^{0} x e^{- 6 x} d x

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