question find the total area between the grap...

# Explanation: ## Step1: Split the absolute - value function The function \(y = |x - 1|-1\) can be written as a piece - wise function: \(y=\begin{cases}(x - 1)-1=x - 2, &x\geq1\\-(x - 1)-1=-x, &x<1\end{cases}\). We need to split the integral over the interval \([-5,3]\) at \(x = 1\). ## Step2: Calculate the integral over sub - intervals We calculate the integral \(\int_{-5}^{3}| |x - 1|-1|dx=\int_{-5}^{0}(-(-x))dx+\int_{0}^{1}-(-x)dx+\int_{1}^{3}(x - 2)dx\). First integral: \(\int_{-5}^{0}xdx=\left[\frac{x^{2}}{2}\right]_{-5}^{0}=0-\frac{(-5)^{2}}{2}=-\frac{25}{2}\), but we take the absolute value \(\left|\int_{-5}^{0}xdx\right|=\frac{25}{2}\). Second integral: \(\int_{0}^{1}xdx=\left[\frac{x^{2}}{2}\right]_{0}^{1}=\frac{1}{2}\). Third integral: \(\int_{1}^{3}(x - 2)dx=\left[\frac{x^{2}}{2}-2x\right]_{1}^{3}=(\frac{3^{2}}{2}-2\times3)-(\frac{1^{2}}{2}-2\times1)=(\frac{9}{2}-6)-(\frac{1}{2}-2)=\frac{9 - 12}{2}-\frac{1 - 4}{2}=-\frac{3}{2}+\frac{3}{2}=0\). ## Step3: Sum up the absolute values of the integrals The total area \(A=\frac{25}{2}+\frac{1}{2}+0 = 13\). # Answer: 13