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

# Explanation: ## Step1: Analyze the absolute - value function The function \(y = |x - 1|-2\) can be written as a piece - wise function: \(y=\begin{cases}x - 1-2=x - 3, &x\geq1\\-(x - 1)-2=-x - 1, &x<1\end{cases}\). ## Step2: Find the x - intercepts Set \(y = 0\). For \(y=x - 3\), \(x=3\) (when \(x\geq1\)); for \(y=-x - 1\), \(x=-1\) (when \(x<1\)). ## Step3: Split the integral based on x - intercepts We split the interval \([-4,6]\) into sub - intervals \([-4,-1]\), \([-1,3]\), and \([3,6]\) based on the x - intercepts \(x=-1\) and \(x = 3\). The area \(A=\int_{-4}^{-1}(-(-x - 1))dx+\int_{-1}^{3}(-(|x - 1|-2))dx+\int_{3}^{6}(|x - 1|-2)dx\). For \(\int_{-4}^{-1}(-(-x - 1))dx=\int_{-4}^{-1}(x + 1)dx\). Using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)\), we have \(\left[\frac{x^{2}}{2}+x\right]_{-4}^{-1}=\left(\frac{(-1)^{2}}{2}-1\right)-\left(\frac{(-4)^{2}}{2}-4\right)=\left(\frac{1}{2}-1\right)-\left(8 - 4\right)=-\frac{1}{2}-4=-\frac{9}{2}\), and the absolute value is \(\frac{9}{2}\). For \(\int_{-1}^{3}(-(|x - 1|-2))dx\). When \(x\in[-1,1]\), \(y=-(-x - 1 + 2)=x - 1\); when \(x\in[1,3]\), \(y=-(x - 1-2)=-x + 3\). \(\int_{-1}^{1}(x - 1)dx+\int_{1}^{3}(-x + 3)dx=\left[\frac{x^{2}}{2}-x\right]_{-1}^{1}+\left[-\frac{x^{2}}{2}+3x\right]_{1}^{3}=\left(\frac{1}{2}-1\right)-\left(\frac{1}{2}+1\right)+\left(-\frac{9}{2}+9\right)-\left(-\frac{1}{2}+3\right)=-2 + 2=2\). For \(\int_{3}^{6}(x - 3)dx=\left[\frac{x^{2}}{2}-3x\right]_{3}^{6}=\left(\frac{6^{2}}{2}-18\right)-\left(\frac{3^{2}}{2}-9\right)=0-\left(\frac{9}{2}-9\right)=\frac{9}{2}\). ## Step4: Sum up the areas of sub - regions \(A=\frac{9}{2}+2+\frac{9}{2}=11\). # Answer: 11