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Q. The value of $ \int\limits_0^4 |x-1|dx $ is

KCETKCET 2011Integrals

Solution:

$\int\limits_{0}^{4}|x-1| d x =\int\limits_{0}^{1}-(x-1) d x+\int\limits_{1}^{4}(x-1) d x$
$=-\left[\frac{x^{2}}{2}-x\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{4}$
$=-[1 / 2-1]+[8-4-1 / 2+1]$
$=- (-1 / 2)+(4+1 / 2)$
$=1 / 2+9 / 2$
$=5$