Question

If $f\left(x\right)=min\{|x-1|,|x-2|,|x-3|\},$ then the value of $I=\underset{0}{\overset{3}{\int }}f\left(x\right)dx$ equals

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $4$

Answer 1

The correct option is A $1$

$f\left(x\right)=min\{|x-1|,|x-2|,|x-3|\}$


From above graph,
$I=\underset{0}{\overset{3}{\int }}f\left(x\right)dx=\frac{1}{2}\cdot 1\cdot 1+\frac{1}{2}\cdot 1\cdot \frac{1}{2}+\frac{1}{2}\cdot 1\cdot \frac{1}{2}=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1$