int_{0}^{2} |x - 1| , dx = | vedclass

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(D) Let $I = \int_{0}^{2} |x - 1| \, dx$. Since the function $|x - 1|$ changes its sign at $x = 1$,we split the integral at $x = 1$: $I = \int_{0}^{1}

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$1$

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(D) Let $I = \int_{0}^{2} |x - 1| \, dx$. Since the function $|x - 1|$ changes its sign at $x = 1$,we split the integral at $x = 1$: $I = \int_{0}^{1} -(x - 1) \, dx + \int_{1}^{2} (x - 1) \, dx$. Evaluating the first integral: $\int_{0}^{1} (-x + 1) \, dx = \left[ -\frac{x^2}{2} + x \right]_{0}^{1} = (-\frac{1}{2} + 1) - (0) = \frac{1}{2}$. Evaluating the second integral: $\int_{1}^{2} (x - 1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{2} = (\frac{4}{2} - 2) - (\frac{1}{2} - 1) = (2 - 2) - (-\frac{1}{2}) = \frac{1}{2}$. Adding both parts: $I = \frac{1}{2} + \frac{1}{2} = 1$.

$\int_{0}^{2} |x - 1| \, dx = $

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