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

解

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

\frac{7}{6}

十進法表記

1.16666 \dots

解答ステップ

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

拡張

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

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

\int_{1}^{4} x d x = \frac{15}{2}

\int_{1}^{4} 2 x d x = \frac{28}{3}

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

= \frac{15}{2} - \frac{28}{3} + 3

簡素化

\frac{15}{2} - \frac{28}{3} + 3 : \frac{7}{6}

= \frac{7}{6}