Tính $\int\limits_{0}^{2}{\sqrt{{{x}^{2}}-2x+1}} \text{d}x$.

Ta có $\int\limits_{0}^{2}{\sqrt{{{x}^{2}}-2x+1}} \text{d}x=\int\limits_{0}^{2}{\sqrt{{{\left( x-1 \right)}^{2}}}}\text{d}x=\int\limits_{0}^{2}{\left| x-1 \right|}\text{d}x=\int\limits_{0}^{1}{-\left( x-1 \right)}\text{d}x+\int\limits_{1}^{2}{\left( x-1 \right)}\text{d}x$
$=-\left. \left( \dfrac{{{x}^{2}}}{2}-x \right) \right|_{0}^{1}+\left. \left( \dfrac{{{x}^{2}}}{2}-x \right) \right|_{1}^{2}=\dfrac{1}{2}+\dfrac{1}{2}=1$.