intlimits_{frac{1}{2}}^{3frac{1}{2}} {left{ { | vedclass

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(C) Let $f(x) = \frac{1}{2}(|x - 3| + |1 - x| - 4)$.For $x \in [\frac{1}{2}, 1]$,$f(x) = \frac{1}{2}(3 - x + 1 - x - 4) = \frac{1}{2}(-2x) = -x$. Thus

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$\frac{1}{4}$

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(C) Let $f(x) = \frac{1}{2}(|x - 3| + |1 - x| - 4)$.For $x \in [\frac{1}{2}, 1]$,$f(x) = \frac{1}{2}(3 - x + 1 - x - 4) = \frac{1}{2}(-2x) = -x$. Thus,$\{f(x)\} = \{-x\} = 1 - x$ (since $x \in [\frac{1}{2}, 1]$,$-x \in [-1, -\frac{1}{2}]$,so $\{-x\} = -x - (-1) = 1 - x$).For $x \in [1, 3]$,$f(x) = \frac{1}{2}(3 - x + x - 1 - 4) = \frac{1}{2}(-2) = -1$. Thus,$\{f(x)\} = \{-1\} = 0$.For $x \in [3, \frac{7}{2}]$,$f(x) = \frac{1}{2}(x - 3 + x - 1 - 4) = \frac{1}{2}(2x - 8) = x - 4$. Thus,$\{f(x)\} = \{x - 4\} = x - 4$ (since $x \in [3, \frac{7}{2}]$,$x - 4 \in [-1, -\frac{1}{2}]$,so $\{x - 4\} = x - 4 - (-1) = x - 3$).Now,the integral is $\int_{\frac{1}{2}}^{1} (1 - x) dx + \int_{1}^{3} 0 dx + \int_{3}^{\frac{7}{2}} (x - 3) dx$.$= [x - \frac{x^2}{2}]_{\frac{1}{2}}^{1} + 0 + [\frac{x^2}{2} - 3x]_{3}^{\frac{7}{2}}$.$= (1 - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{8}) + ((\frac{49}{8} - \frac{21}{2}) - (\frac{9}{2} - 9))$.$= \frac{1}{2} - \frac{3}{8} + (\frac{49 - 84}{8} - (-\frac{9}{2})) = \frac{1}{8} + (-\frac{35}{8} + \frac{36}{8}) = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$.

$\int\limits_{\frac{1}{2}}^{3\frac{1}{2}} {\left\{ {\frac{1}{2}\,\left( {|x - 3| + |1 - x| - 4} \right)} \right\}\,dx} $ equals: Where $\{*\}$ denotes the fractional part function.

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