[IIT Screening]
A) 10
B) $ \frac{5}{6} $
C) 21
D) 12
Correct Answer: D
$ \int_1^{5}{(|x-3|+|1-x|)dx} $
$ =\int_1^{5}{|x-3|dx+\int_1^{5}{|1-x|}dx} $
$ =\int_1^{3}{-(x-3)dx+\int_3^{5}{(x-3)}dx+}\int_1^{5}{-(1-x)}dx $
$ =12 $ .
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