question find the total area between the grap...

# Explanation: ## Step1: Split 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: Calculate the area using integration We split the integral over the interval \([-3,3]\) into two parts: \(\int_{-3}^{1}x dx+\int_{1}^{3}(2 - x)dx\). First, calculate \(\int_{-3}^{1}x dx\). Using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)\), we have \(\left[\frac{x^{2}}{2}\right]_{-3}^{1}=\frac{1^{2}}{2}-\frac{(-3)^{2}}{2}=\frac{1}{2}-\frac{9}{2}=-4\). But we want the area, so we take the absolute - value \(\vert - 4\vert = 4\). Second, calculate \(\int_{1}^{3}(2 - x)dx=\left[2x-\frac{x^{2}}{2}\right]_{1}^{3}=(2\times3-\frac{3^{2}}{2})-(2\times1-\frac{1^{2}}{2})=(6-\frac{9}{2})-(2 - \frac{1}{2})=( \frac{12 - 9}{2})-( \frac{4 - 1}{2})=\frac{3}{2}-\frac{3}{2}=0\). The total area \(A = 4+0 = 4\). # Answer: 4