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# Explanation: ## Step1: Rewrite the absolute - value function For \(y = |x - 2|-1\), when \(x\geq2\), \(y=(x - 2)-1=x - 3\); when \(x<2\), \(y=-(x - 2)-1=-x + 1\). ## Step2: Split the integral based on the break - point The break - point of \(y = |x - 2|-1\) is \(x = 2\). We split the integral \(\int_{-2}^{6}| |x - 2|-1|dx\) into three parts: \(\int_{-2}^{1}(-(-x + 1))dx+\int_{1}^{2}(-x + 1)dx+\int_{2}^{5}(x - 3)dx+\int_{5}^{6}(x - 3)dx\). First, for \(\int_{-2}^{1}(-(-x + 1))dx=\int_{-2}^{1}(x - 1)dx\). Using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)\), we have \(\left[\frac{x^{2}}{2}-x\right]_{-2}^{1}=\left(\frac{1}{2}-1\right)-\left(\frac{4}{2}+2\right)=\frac{1 - 2}{2}-(2 + 2)=-\frac{1}{2}-4=-\frac{9}{2}\), and its absolute value is \(\frac{9}{2}\). Second, \(\int_{1}^{2}(-x + 1)dx=\left[-\frac{x^{2}}{2}+x\right]_{1}^{2}=\left(-\frac{4}{2}+2\right)-\left(-\frac{1}{2}+1\right)=0-\frac{1}{2}=-\frac{1}{2}\), and its absolute value is \(\frac{1}{2}\). Third, \(\int_{2}^{5}(x - 3)dx=\left[\frac{x^{2}}{2}-3x\right]_{2}^{5}=\left(\frac{25}{2}-15\right)-\left(\frac{4}{2}-6\right)=\frac{25 - 30}{2}-\left(2 - 6\right)=-\frac{5}{2}+4=\frac{3}{2}\), and its absolute value is \(\frac{3}{2}\). Fourth, \(\int_{5}^{6}(x - 3)dx=\left[\frac{x^{2}}{2}-3x\right]_{5}^{6}=\left(\frac{36}{2}-18\right)-\left(\frac{25}{2}-15\right)=0-\left(\frac{25 - 30}{2}\right)=\frac{5}{2}\), and its absolute value is \(\frac{5}{2}\). ## Step3: Sum up the absolute - value of the integrals \(A=\frac{9}{2}+\frac{1}{2}+\frac{3}{2}+\frac{5}{2}=\frac{9 + 1+3 + 5}{2}=9\). # Answer: 9