integral from 0 to 2 of 1-c|1-x|

解

\int_{0}^{2} 1 - c ∣ 1 - x ∣ d x

- c + 2

解答ステップ

\int_{0}^{2} 1 - c ∣ 1 - x ∣ d x

絶対数を排除する

= \int_{0}^{1} 1 - c (1 - x) d x + \int_{1}^{2} 1 - c (- 1 + x) d x

\int_{0}^{1} 1 - c (1 - x) d x = 1 - \frac{1}{2} c

\int_{1}^{2} 1 - c (- 1 + x) d x = 1 - \frac{1}{2} c

= 1 - \frac{1}{2} c + 1 - \frac{1}{2} c

簡素化

1 - \frac{1}{2} c + 1 - \frac{1}{2} c : - c + 2

= - c + 2