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

解

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

\frac{256}{15}

十進法表記

17.06666 \dots

解答ステップ

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

拡張

(4 - x^{2})^{2} : 16 - 8 x^{2} + x^{4}

= \int_{0}^{2} 16 - 8 x^{2} + x^{4} 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}^{2} 16 d x - \int_{0}^{2} 8 x^{2} d x + \int_{0}^{2} x^{4} d x

\int_{0}^{2} 16 d x = 32

\int_{0}^{2} 8 x^{2} d x = \frac{64}{3}

\int_{0}^{2} x^{4} d x = \frac{32}{5}

= 32 - \frac{64}{3} + \frac{32}{5}

簡素化

32 - \frac{64}{3} + \frac{32}{5} : \frac{256}{15}

= \frac{256}{15}