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

解

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

\frac{8}{3}

十進法表記

2.66666 \dots

解答ステップ

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

拡張

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

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

\int_{0}^{4} 4 d x = 16

\int_{0}^{4} 4 x d x = \frac{64}{3}

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

= 16 - \frac{64}{3} + 8

簡素化

16 - \frac{64}{3} + 8 : \frac{8}{3}

= \frac{8}{3}