integral from 0 to 2pi of x^2sin(2x)

解答

\int_{0}^{2 π} x^{2} sin (2 x) d x

- 2 π^{2}

十进制

- 19.73920 \dots

求解步骤

\int_{0}^{2 π} x^{2} sin (2 x) d x

使用分布积分法

= [\frac{1}{2} (- x^{2} cos (2 x) - 2 \cdot \int - x cos (2 x) d x)]_{0}^{2 π}

\int - x cos (2 x) d x = - \frac{1}{4} (2 x sin (2 x) + cos (2 x))

= [\frac{1}{2} (- x^{2} cos (2 x) - 2 (- \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{2 π}

化简

[\frac{1}{2} (- x^{2} cos (2 x) - 2 (- \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{2 π} : [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{2 π}

= [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{2 π}

带入上下限计算后得:

\frac{- 8 π ^{2} + 1}{4} - \frac{1}{4}

= \frac{- 8 π ^{2} + 1}{4} - \frac{1}{4}

化简

= - 2 π^{2}

\frac{d}{d t} (3 sin (3 t))

(\frac{2 y}{x} + 2 x) + (2 ln (x) - 3) y^{^{'}} = 0

\frac{\partial}{\partial x} (\frac{26 x ^{2}}{( x - y ) ^{2}})

t an g e n t 1 + 4 x^{2} - 2 x^{3}

\frac{d}{d x} (sin (x - 1) (x^{3}))