The value of the integral int_{1}^{5}[|x-3|+| | vedclass

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(C) Let $I = \int_{1}^{5} [|x-3| + |1-x|] dx$. Since $x \in [1, 5]$,$|1-x| = x-1$. Thus,$I = \int_{1}^{5} |x-3| dx + \int_{1}^{5} (x-1) dx$. For the f

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$12$

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(C) Let $I = \int_{1}^{5} [|x-3| + |1-x|] dx$. Since $x \in [1, 5]$,$|1-x| = x-1$. Thus,$I = \int_{1}^{5} |x-3| dx + \int_{1}^{5} (x-1) dx$. For the first part,$\int_{1}^{5} |x-3| dx = \int_{1}^{3} -(x-3) dx + \int_{3}^{5} (x-3) dx$. $= \int_{1}^{3} (3-x) dx + \int_{3}^{5} (x-3) dx = [3x - \frac{x^2}{2}]_{1}^{3} + [\frac{x^2}{2} - 3x]_{3}^{5}$. $= (9 - 4.5) - (3 - 0.5) + (12.5 - 15) - (4.5 - 9) = 4.5 - 2.5 - 2.5 + 4.5 = 4$. For the second part,$\int_{1}^{5} (x-1) dx = [\frac{x^2}{2} - x]_{1}^{5} = (12.5 - 5) - (0.5 - 1) = 7.5 - (-0.5) = 8$. Therefore,$I = 4 + 8 = 12$.

The value of the integral $\int_{1}^{5}[|x-3|+|1-x|] dx$ is equal to

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