integral from 0 to 1 of x^3(1-3x^4)^2

解

\int_{0}^{1} x^{3} (1 - 3 x^{4})^{2} d x

\frac{1}{4}

十進法表記

0.25

解答ステップ

\int_{0}^{1} x^{3} (1 - 3 x^{4})^{2} d x

拡張

x^{3} (1 - 3 x^{4})^{2} : x^{3} - 6 x^{7} + 9 x^{11}

= \int_{0}^{1} x^{3} - 6 x^{7} + 9 x^{11} 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^{3} d x - \int_{0}^{1} 6 x^{7} d x + \int_{0}^{1} 9 x^{11} d x

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} 6 x^{7} d x = \frac{3}{4}

\int_{0}^{1} 9 x^{11} d x = \frac{3}{4}

= \frac{1}{4} - \frac{3}{4} + \frac{3}{4}

簡素化

= \frac{1}{4}