The value of the integral $\int\limits_1^3|x-2| d x$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - 1
$\int\limits_1^3|x-2| d x$
$=\int\limits_1^2(2-x)dx+\int\limits_2^3x-2dx$
$=\left|2x-\frac{x^2}{2}\right|_1^2+\left|\frac{x^2}{2}-2x\right|_2^3$
$=4-\frac{4}{2}-2+\frac{1}{2}+\frac{9}{2}-6-\frac{4}{2}+4$
$=1$