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

解答

\int_{0}^{2 π} x sin^{2} (x) d x

π^{2}

十进制

9.86960 \dots

求解步骤

\int_{0}^{2 π} x sin^{2} (x) d x

使用分布积分法

= [\frac{1}{4} (x (2 x - sin (2 x)) - 4 \cdot \int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x)]_{0}^{2 π}

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

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

化简

[\frac{1}{4} (x (2 x - sin (2 x)) - 4 \cdot \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x)))]_{0}^{2 π} : [\frac{1}{8} (2 x^{2} - 2 x sin (2 x) - cos (2 x))]_{0}^{2 π}

= [\frac{1}{8} (2 x^{2} - 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}{8} + \frac{1}{8}

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

化简

= π^{2}

\int_{0}^{4} (- x^{2} - 3 x) d x

\int_{0}^{3} x^{2} + 5 d x

\int_{- 1}^{3} 8 x^{3} d x

\int_{a}^{2 a} (a + x) d x

\int_{1}^{5} x^{2} - 2 x - 8 d x