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

解

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

\frac{28}{3}

十進法表記

9.33333 \dots

解答ステップ

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

拡張

(x - 1)^{2} : x^{2} - 2 x + 1

= \int_{0}^{4} x^{2} - 2 x + 1 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} x^{2} d x - \int_{0}^{4} 2 x d x + \int_{0}^{4} 1 d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

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

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

= \frac{64}{3} - 16 + 4

簡素化

\frac{64}{3} - 16 + 4 : \frac{28}{3}

= \frac{28}{3}