integral from 0 to 6pi of x^2sin(2x)

解答

\int_{0}^{6 π} x^{2} sin (2 x) d x

- 18 π^{2}

十进制

- 177.65287 \dots

求解步骤

\int_{0}^{6 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}^{6 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}^{6 π}

化简

[\frac{1}{2} (- x^{2} cos (2 x) - 2 (- \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{6 π} : [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{6 π}

= [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{6 π}

带入上下限计算后得:

\frac{- 72 π ^{2} + 1}{4} - \frac{1}{4}

= \frac{- 72 π ^{2} + 1}{4} - \frac{1}{4}

化简

= - 18 π^{2}

d er i v a t i v e ln (x^{2} - 4 x + 1)

\frac{\partial}{\partial y} (x \cdot y^{2})

x \to 0 lim (\frac{( 3 x + 8 - 2 )}{x})

\int 5 x \cdot cos (8 x) d x

\frac{d}{d x} (- sin^{2} (x))