If f(x)=|x-1|+|x-2|+|x-3|, forall x in[1,4],t | vedclass

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(D) We need to evaluate the integral $I = \int_1^4 (|x-1|+|x-2|+|x-3|) dx$.Since the absolute value functions change their behavior at $x=1, 2, 3$,we

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$\frac{19}{2}$

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(D) We need to evaluate the integral $I = \int_1^4 (|x-1|+|x-2|+|x-3|) dx$.Since the absolute value functions change their behavior at $x=1, 2, 3$,we split the integral into three intervals:$I = \int_1^2 ((x-1) + (2-x) + (3-x)) dx + \int_2^3 ((x-1) + (x-2) + (3-x)) dx + \int_3^4 ((x-1) + (x-2) + (x-3)) dx$Simplifying the integrands:$I = \int_1^2 (4-x) dx + \int_2^3 x dx + \int_3^4 (3x-6) dx$Now,integrate each part:$\int_1^2 (4-x) dx = [4x - \frac{x^2}{2}]_1^2 = (8-2) - (4-0.5) = 6 - 3.5 = 2.5$$\int_2^3 x dx = [\frac{x^2}{2}]_2^3 = \frac{9}{2} - \frac{4}{2} = 4.5 - 2 = 2.5$$\int_3^4 (3x-6) dx = [\frac{3x^2}{2} - 6x]_3^4 = (24-24) - (13.5-18) = 0 - (-4.5) = 4.5$Adding these values: $I = 2.5 + 2.5 + 4.5 = 9.5 = \frac{19}{2}$.

If $f(x)=|x-1|+|x-2|+|x-3|, \forall x \in[1,4]$,then $\int_1^4 f(x) dx=$

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