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

解

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

\frac{512}{15}

十進法表記

34.13333 \dots

解答ステップ

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

拡張

(4 x - x^{2})^{2} : 16 x^{2} - 8 x^{3} + x^{4}

= \int_{0}^{4} 16 x^{2} - 8 x^{3} + 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}^{4} 16 x^{2} d x - \int_{0}^{4} 8 x^{3} d x + \int_{0}^{4} x^{4} d x

\int_{0}^{4} 16 x^{2} d x = \frac{1024}{3}

\int_{0}^{4} 8 x^{3} d x = 512

\int_{0}^{4} x^{4} d x = \frac{1024}{5}

= \frac{1024}{3} - 512 + \frac{1024}{5}

簡素化

\frac{1024}{3} - 512 + \frac{1024}{5} : \frac{512}{15}

= \frac{512}{15}