find the total area between the graph of the ...

# Explanation: ## Step1: Analyze the absolute - value function First, rewrite \(y = |x - 3|-2\) as a piece - wise function. When \(x-3\geq0\) (i.e., \(x\geq3\)), \(y=(x - 3)-2=x - 5\). When \(x-3<0\) (i.e., \(x<3\)), \(y=-(x - 3)-2=-x + 1\). ## Step2: Find the x - intercepts For \(y=-x + 1\), set \(y = 0\), then \(0=-x + 1\), so \(x = 1\). For \(y=x - 5\), set \(y = 0\), then \(0=x - 5\), so \(x = 5\). ## Step3: Split the integral based on the intervals and x - intercepts We split the interval \([-6,6]\) into sub - intervals \([-6,1]\), \([1,3]\), \([3,5]\) and \([5,6]\). The area \(A=\int_{-6}^{1}(-(-x + 1))dx+\int_{1}^{3}(-x + 1)dx+\int_{3}^{5}(x - 5)dx+\int_{5}^{6}(x - 5)dx\). ## Step4: Calculate each integral For \(\int_{-6}^{1}(x - 1)dx=\left[\frac{x^{2}}{2}-x\right]_{-6}^{1}=\left(\frac{1^{2}}{2}-1\right)-\left(\frac{(-6)^{2}}{2}+6\right)=\frac{1}{2}-1-(18 + 6)=\frac{1 - 2}{2}-24=-\frac{1}{2}-24=-\frac{49}{2}\), and the absolute value is \(\frac{49}{2}\). For \(\int_{1}^{3}(-x + 1)dx=\left[-\frac{x^{2}}{2}+x\right]_{1}^{3}=\left(-\frac{3^{2}}{2}+3\right)-\left(-\frac{1^{2}}{2}+1\right)=\left(-\frac{9}{2}+3\right)-\left(-\frac{1}{2}+1\right)=-\frac{3}{2}-\frac{1}{2}=-2\), and the absolute value is \(2\). For \(\int_{3}^{5}(x - 5)dx=\left[\frac{x^{2}}{2}-5x\right]_{3}^{5}=\left(\frac{5^{2}}{2}-25\right)-\left(\frac{3^{2}}{2}-15\right)=\frac{25}{2}-25-\frac{9}{2}+15=\frac{25 - 9}{2}-10=8 - 10=-2\), and the absolute value is \(2\). For \(\int_{5}^{6}(x - 5)dx=\left[\frac{x^{2}}{2}-5x\right]_{5}^{6}=\left(\frac{6^{2}}{2}-30\right)-\left(\frac{5^{2}}{2}-25\right)=\left(18-30\right)-\left(\frac{25}{2}-25\right)=-12+\frac{25}{2}=\frac{1}{2}\), and the absolute value is \(\frac{1}{2}\). ## Step5: Sum up the areas of sub - intervals \(A=\frac{49}{2}+2 + 2+\frac{1}{2}=\frac{49 + 1}{2}+4=25+4 = 29\). # Answer: \(29\)