integral from-1 to 2 of x^3(x^2+1)^8

解答

\int_{- 1}^{2} x^{3} (x^{2} + 1)^{8} d x

\frac{7595031}{20}

十进制

379751.55

求解步骤

\int_{- 1}^{2} x^{3} (x^{2} + 1)^{8} d x

乘开

x^{3} (x^{2} + 1)^{8} : x^{19} + 8 x^{17} + 28 x^{15} + 56 x^{13} + 70 x^{11} + 56 x^{9} + 28 x^{7} + 8 x^{5} + x^{3}

= \int_{- 1}^{2} x^{19} + 8 x^{17} + 28 x^{15} + 56 x^{13} + 70 x^{11} + 56 x^{9} + 28 x^{7} + 8 x^{5} + x^{3} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{2} x^{19} d x + \int_{- 1}^{2} 8 x^{17} d x + \int_{- 1}^{2} 28 x^{15} d x + \int_{- 1}^{2} 56 x^{13} d x + \int_{- 1}^{2} 70 x^{11} d x + \int_{- 1}^{2} 56 x^{9} d x + \int_{- 1}^{2} 28 x^{7} d x + \int_{- 1}^{2} 8 x^{5} d x + \int_{- 1}^{2} x^{3} d x

\int_{- 1}^{2} x^{19} d x = \frac{209715}{4}

\int_{- 1}^{2} 8 x^{17} d x = 116508

\int_{- 1}^{2} 28 x^{15} d x = \frac{458745}{4}

\int_{- 1}^{2} 56 x^{13} d x = 65532

\int_{- 1}^{2} 70 x^{11} d x = \frac{47775}{2}

\int_{- 1}^{2} 56 x^{9} d x = \frac{28644}{5}

\int_{- 1}^{2} 28 x^{7} d x = \frac{1785}{2}

\int_{- 1}^{2} 8 x^{5} d x = 84

\int_{- 1}^{2} x^{3} d x = \frac{15}{4}

= \frac{209715}{4} + 116508 + \frac{458745}{4} + 65532 + \frac{47775}{2} + \frac{28644}{5} + \frac{1785}{2} + 84 + \frac{15}{4}

化简

\frac{209715}{4} + 116508 + \frac{458745}{4} + 65532 + \frac{47775}{2} + \frac{28644}{5} + \frac{1785}{2} + 84 + \frac{15}{4} : \frac{7595031}{20}

= \frac{7595031}{20}

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