The integral int_{0}^{2} ||x-1|-x| dx is equa | vedclass

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(A) Let $f(x) = ||x-1|-x|$.We analyze the expression inside the absolute value:If $x \geq 1$,then $|x-1| = x-1$,so $f(x) = |(x-1)-x| = |-1| = 1$.If $x

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(A) Let $f(x) = ||x-1|-x|$.We analyze the expression inside the absolute value:If $x \geq 1$,then $|x-1| = x-1$,so $f(x) = |(x-1)-x| = |-1| = 1$.If $x Thus,$f(x) = \begin{cases} |1-2x|, & 0 \leq x Now,we evaluate the integral:$\int_{0}^{2} f(x) dx = \int_{0}^{1} |1-2x| dx + \int_{1}^{2} 1 dx$.For the first part,$|1-2x| = 1-2x$ when $x \leq 1/2$ and $2x-1$ when $x > 1/2$:$\int_{0}^{1} |1-2x| dx = \int_{0}^{1/2} (1-2x) dx + \int_{1/2}^{1} (2x-1) dx$$= [x-x^2]_{0}^{1/2} + [x^2-x]_{1/2}^{1}$$= (1/2 - 1/4) - 0 + (1-1) - (1/4 - 1/2) = 1/4 + 1/4 = 1/2$.For the second part:$\int_{1}^{2} 1 dx = [x]_{1}^{2} = 2-1 = 1$.Total integral = $1/2 + 1 = 1.5$.

The integral $\int_{0}^{2} ||x-1|-x| dx$ is equal to (in $.5$)

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