\(\displaystyle \int \limits_0^2 \sqrt{x^2-2x+1}   dx =\)

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Integrate by completing the square.

$$\int \limits_0^2 \sqrt{x^2-2x+1}   dx = \int \limits_0^2 \sqrt{(x-1)^2}   dx = \int \limits_0^2 x-1   dx $$ $$ = \left[\frac{1}{2}x^2-x\right]_0^2 $$ $$ =\left(\frac{1}{2}(2)^2-2\right)-\left(\frac{1}{2}(0)^2-0\right) $$ $$ =(2-2)-(0-0) $$ $$ =\boxed{0} $$