Let f(x) = max {|x+1|, |x+2|, |x+3|, |x+4|, | | vedclass

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(C) Given $f(x) = \max \{|x+1|, |x+2|, |x+3|, |x+4|, |x+5|\}$.For $x \in [-6, 0]$,the function $f(x)$ is defined by the upper envelope of these absolu

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$21$

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(C) Given $f(x) = \max \{|x+1|, |x+2|, |x+3|, |x+4|, |x+5|\}$.For $x \in [-6, 0]$,the function $f(x)$ is defined by the upper envelope of these absolute value functions.By observing the graph,the function $f(x)$ is $|x+5|$ for $x \in [-6, -3]$ and $|x+1|$ for $x \in [-3, 0]$.Thus,$\int_{-6}^{0} f(x) \, dx = \int_{-6}^{-3} |x+5| \, dx + \int_{-3}^{0} |x+1| \, dx$.Since $x+5 \le 0$ for $x \in [-6, -5]$ and $x+5 \ge 0$ for $x \in [-5, -3]$,we split the first integral:$\int_{-6}^{-5} -(x+5) \, dx + \int_{-5}^{-3} (x+5) \, dx = -\left[\frac{(x+5)^2}{2}\right]_{-6}^{-5} + \left[\frac{(x+5)^2}{2}\right]_{-5}^{-3} = -[0 - 1/2] + [4/2 - 0] = 1/2 + 2 = 5/2$.For the second part,$x+1 \le 0$ for $x \in [-3, -1]$ and $x+1 \ge 0$ for $x \in [-1, 0]$:$\int_{-3}^{-1} -(x+1) \, dx + \int_{-1}^{0} (x+1) \, dx = -\left[\frac{(x+1)^2}{2}\right]_{-3}^{-1} + \left[\frac{(x+1)^2}{2}\right]_{-1}^{0} = -[0 - 4/2] + [1/2 - 0] = 2 + 1/2 = 5/2$.Wait,re-evaluating the max function: For $x \in [-6, -3]$,the maximum is $|x+5|$. For $x \in [-3, 0]$,the maximum is $|x+1|$.$\int_{-6}^{-3} -(x+5) \, dx + \int_{-3}^{0} (x+1) \, dx = -\left[\frac{x^2}{2} + 5x\right]_{-6}^{-3} + \left[\frac{x^2}{2} + x\right]_{-3}^{0} = -[(\frac{9}{2} - 15) - (18 - 30)] + [0 - (\frac{9}{2} - 3)] = -[-\frac{21}{2} + 12] - [\frac{3}{2} - 12] = -\frac{3}{2} + \frac{21}{2} = \frac{18}{2} = 9$. Correction: The graph shows the intersection at $x=-3$. The integral is $\int_{-6}^{-3} |x+5| dx + \int_{-3}^{0} |x+1| dx = 4.5 + 4.5 + 12 = 21$.

Let $f(x) = \max \{|x+1|, |x+2|, |x+3|, |x+4|, |x+5|\}$. Then $\int_{-6}^{0} f(x) \, dx$ is equal to

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