If $f(x)=Min\{|x-1|,|x|,|x+1|\}$, then $\int\limits_{-1}^1 f(x) d x$ equals
Answer & explanation
Correct answer: option 4
The graph of f(x) is as shown in figure.
∴ $\int\limits_{-1}^1 f(x) d x=2 \int\limits_0^1 f(x) d x=2\left(\frac{1}{2} \times 1 \times \frac{1}{2}\right)=\frac{1}{2}$
