Question

The value of the integral ${\int }_{-2}^{2}min.\{|x-1|,|x+2|\}dx$
is

• A. $\frac{11}{4}$
  
• B. $\frac{9}{4}$
  
• C. $\frac{11}{2}$
  
• D. $\frac{9}{2}$

Answer 1

The correct option is A $\frac{11}{4}$


$f\left(x\right)=\{\begin{array}{ccc}x+2& if& x<\frac{-1}{2}\\ 1-x& if& \frac{-1}{2}<x<1\\ x-1& if& x>1\end{array}$
$\therefore {\int }_{-2}^{2}f\left(x\right)dx={\int }_{-2}^{-1/2}x+2+{\int }_{-1/2}^{1}1-x+{\int }_1^{2}x-1=\frac{11}{4}$