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

解

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

\frac{19}{6}

十進法表記

3.16666 \dots

解答ステップ

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

絶対数を排除する

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

\int_{1}^{4} x d x = \frac{14}{3}

\int_{1}^{4} 2 d x = 6

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

= \frac{14}{3} + 6 - \frac{15}{2}

簡素化

\frac{14}{3} + 6 - \frac{15}{2} : \frac{19}{6}

= \frac{19}{6}