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$\int_{0}^{4}|x-1| d x$


$I=\int_{0}^{4}|x-1| d x$

It can be seen that, $(x-1) \leq 0$ when $0 \leq x \leq 1$ and $(x-1) \geq 0$ when $1 \leq x \leq 4$

$I=\int_{1}^{1}|x-1| d x+\int_{1}^{4}|x-1| d x$                     $\left(\int_{a}^{b} f(x)=\int_{a}^{c} f(x)+\int_{c}^{b} f(x)\right)$

$=\int_{0}^{1}-(x-1) d x+\int_{0}^{4}(x-1) d x$





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