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

# Explanation: ## Step1: Analyze the absolute - value function The function \(y = 1-\vert x - 1\vert\) can be written as a piece - wise function. When \(x-1\geq0\) (i.e., \(x\geq1\)), \(y = 1-(x - 1)=2 - x\); when \(x - 1<0\) (i.e., \(x<1\)), \(y=1-(1 - x)=x\). ## Step2: Find the intersection points with the x - axis Set \(y = 0\). For \(y=x\), \(x = 0\); for \(y=2 - x\), \(x = 2\). ## Step3: Split the integral based on intervals We split the interval \([-6,4]\) into sub - intervals based on the behavior of the function and its intersection with the \(x\) - axis. We need to calculate the integral \(\int_{-6}^{4}\vert1-\vert x - 1\vert\vert dx\). We split it into three parts: \(\int_{-6}^{0}-(1-\vert x - 1\vert)dx+\int_{0}^{2}(1-\vert x - 1\vert)dx+\int_{2}^{4}-(1-\vert x - 1\vert)dx\). For \(-6\leq x<0\), \(y=1-(1 - x)=x\), and \(\int_{-6}^{0}-x dx=-\left[\frac{x^{2}}{2}\right]_{-6}^{0}=-\left(0-\frac{(-6)^{2}}{2}\right)=18\). For \(0\leq x<1\), \(y = x\), and \(\int_{0}^{1}x dx=\left[\frac{x^{2}}{2}\right]_{0}^{1}=\frac{1}{2}\). For \(1\leq x<2\), \(y=2 - x\), and \(\int_{1}^{2}(2 - x)dx=\left[2x-\frac{x^{2}}{2}\right]_{1}^{2}=(4 - 2)-(2-\frac{1}{2})=\frac{1}{2}\). For \(2\leq x<4\), \(y=-(2 - x)=x - 2\), and \(\int_{2}^{4}(x - 2)dx=\left[\frac{x^{2}}{2}-2x\right]_{2}^{4}=(\frac{4^{2}}{2}-8)-(\frac{2^{2}}{2}-4)=2\). ## Step4: Sum up the areas The total area \(A=18+\frac{1}{2}+\frac{1}{2}+2 = 21\). # Answer: 21