1) The value of the integral ∫3−1(|x|+|x−1|)dx is A) 4 B) 9 C) 2 D) 9/2 Answer: Option BExplanation:|x|+|x−1|={−x−(x−1)=−2x+1,ifx≤0x−(x−1)=1,if0≤x≤1x+(x−1)=2x−1,ifx≥1 .'.∫3−1(|x|+|x−1|)dx = ∫0−1(−2x+1)dx+∫101dx+∫31(2x−1)dx = [−x2+x]0−1+[x]10+[x2−x]31 =9