integral from 0 to 1 of x^2(x^3+1)^3

解

\int_{0}^{1} x^{2} (x^{3} + 1)^{3} d x

\frac{5}{4}

十進法表記

1.25

解答ステップ

\int_{0}^{1} x^{2} (x^{3} + 1)^{3} d x

拡張

x^{2} (x^{3} + 1)^{3} : x^{11} + 3 x^{8} + 3 x^{5} + x^{2}

= \int_{0}^{1} x^{11} + 3 x^{8} + 3 x^{5} + x^{2} d x

総和規則を適用する:

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

= \int_{0}^{1} x^{11} d x + \int_{0}^{1} 3 x^{8} d x + \int_{0}^{1} 3 x^{5} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} x^{11} d x = \frac{1}{12}

\int_{0}^{1} 3 x^{8} d x = \frac{1}{3}

\int_{0}^{1} 3 x^{5} d x = \frac{1}{2}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= \frac{1}{12} + \frac{1}{3} + \frac{1}{2} + \frac{1}{3}

簡素化

\frac{1}{12} + \frac{1}{3} + \frac{1}{2} + \frac{1}{3} : \frac{5}{4}

= \frac{5}{4}