int_{2}^{4} (|x - 2| + |x - 3|) dx = | vedclass

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(C) We need to evaluate the integral $I = \int_{2}^{4} (|x - 2| + |x - 3|) dx$. Split the integral based on the critical points $x = 2$ and $x = 3$. F

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$3$

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(C) We need to evaluate the integral $I = \int_{2}^{4} (|x - 2| + |x - 3|) dx$. Split the integral based on the critical points $x = 2$ and $x = 3$. For $2 \le x \le 3$,$|x - 2| = x - 2$ and $|x - 3| = 3 - x$. For $3 \le x \le 4$,$|x - 2| = x - 2$ and $|x - 3| = x - 3$. Thus,$I = \int_{2}^{3} ((x - 2) + (3 - x)) dx + \int_{3}^{4} ((x - 2) + (x - 3)) dx$. $I = \int_{2}^{3} (1) dx + \int_{3}^{4} (2x - 5) dx$. $I = [x]_{2}^{3} + [x^2 - 5x]_{3}^{4}$. $I = (3 - 2) + ((16 - 20) - (9 - 15))$. $I = 1 + (-4 - (-6)) = 1 + 2 = 3$.

$\int_{2}^{4} (|x - 2| + |x - 3|) dx =$

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