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By using the properties of definite integrals, evaluate the integrals
$\int_0^4|x-1| d x=\mathrm{I}($ say $)$
$\int_0^4|x-1| d x=\mathrm{I}($ say $)$
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$\mathrm{I}=-\int_0^1(x-1) d x+\int_1^4(x-1) d x$
$=-\left(\frac{x^2}{2}-x\right)_0^1+\left(\frac{x^2}{2}-x\right)_1^4=5$
$=-\left(\frac{x^2}{2}-x\right)_0^1+\left(\frac{x^2}{2}-x\right)_1^4=5$
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