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

# Explanation: ## Step1: Rewrite the absolute - value function The function \(y = |x - 3|-2\) can be rewritten 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 intersection points with the x - axis Set \(y = 0\). For \(y=-x + 1\), when \(y = 0\), \(x = 1\). For \(y=x - 5\), when \(y = 0\), \(x = 5\). ## Step3: Split the integral based on the intervals We need to split the integral \(\int_{-4}^{6}| |x - 3|-2|dx\) into sub - integrals based on the intervals \([-4,1]\), \([1,3]\), \([3,5]\) and \([5,6]\). On \([-4,1]\), \(y=-x + 1\geq0\), the area \(A_1=\int_{-4}^{1}(-x + 1)dx=\left[-\frac{x^{2}}{2}+x\right]_{-4}^{1}=-\frac{1}{2}+1-\left(-\frac{(-4)^{2}}{2}-4\right)=-\frac{1}{2}+1 + 8 + 4=\frac{-1 + 2+16 + 8}{2}=\frac{25}{2}\). On \([1,3]\), \(y=-x + 1\leq0\), the area \(A_2=\int_{1}^{3}(x - 1)dx=\left[\frac{x^{2}}{2}-x\right]_{1}^{3}=\frac{9}{2}-3-\left(\frac{1}{2}-1\right)=\frac{9 - 6-1 + 2}{2}=2\). On \([3,5]\), \(y=x - 5\leq0\), the area \(A_3=\int_{3}^{5}(5 - x)dx=\left[5x-\frac{x^{2}}{2}\right]_{3}^{5}=25-\frac{25}{2}-\left(15-\frac{9}{2}\right)=25-\frac{25}{2}-15+\frac{9}{2}=\frac{50 - 25-30 + 9}{2}=2\). On \([5,6]\), \(y=x - 5\geq0\), the area \(A_4=\int_{5}^{6}(x - 5)dx=\left[\frac{x^{2}}{2}-5x\right]_{5}^{6}=\frac{36}{2}-30-\left(\frac{25}{2}-25\right)=18-30-\frac{25}{2}+25=\frac{36 - 60-25 + 50}{2}=\frac{1}{2}\). ## Step4: Calculate the total area The total area \(A=A_1+A_2+A_3+A_4=\frac{25}{2}+2 + 2+\frac{1}{2}=\frac{25 + 4+4 + 1}{2}=17\). # Answer: 17