If f(x) = |x| + |x - 1| + |x - 2|,x in R,then | vedclass

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(C) To evaluate $\int_{0}^{3} (|x| + |x - 1| + |x - 2|) \, dx$,we split the integral based on the critical points $x = 0, 1, 2$.Since $x \ge 0$ in the

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$19/2$

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(C) To evaluate $\int_{0}^{3} (|x| + |x - 1| + |x - 2|) \, dx$,we split the integral based on the critical points $x = 0, 1, 2$.Since $x \ge 0$ in the interval $[0, 3]$,$|x| = x$.For $x \in [0, 1]$,$|x-1| = 1-x$ and $|x-2| = 2-x$. So $f(x) = x + (1-x) + (2-x) = 3-x$.For $x \in [1, 2]$,$|x-1| = x-1$ and $|x-2| = 2-x$. So $f(x) = x + (x-1) + (2-x) = x+1$.For $x \in [2, 3]$,$|x-1| = x-1$ and $|x-2| = x-2$. So $f(x) = x + (x-1) + (x-2) = 3x-3$.Now,$\int_{0}^{3} f(x) \, dx = \int_{0}^{1} (3-x) \, dx + \int_{1}^{2} (x+1) \, dx + \int_{2}^{3} (3x-3) \, dx$.$= [3x - \frac{x^2}{2}]_{0}^{1} + [\frac{x^2}{2} + x]_{1}^{2} + [\frac{3x^2}{2} - 3x]_{2}^{3}$.$= (3 - 0.5) + (2 + 2 - 0.5 - 1) + (13.5 - 9 - 6 + 6) = 2.5 + 2.5 + 4.5 = 9.5 = 19/2$.

If $f(x) = |x| + |x - 1| + |x - 2|$,$x \in R$,then $\int_{0}^{3} f(x) \, dx = $

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