find the net signed area between the graph of...

# Explanation: ## Step1: Split the absolute - value function The function \(y = 3-\vert x + 3\vert\) can be split into two cases. When \(x+3\geq0\) (i.e., \(x\geq - 3\)), \(y = 3-(x + 3)=-x\). When \(x + 3<0\) (i.e., \(x<-3\)), \(y=3+(x + 3)=x + 6\). ## Step2: Calculate the integral over sub - intervals We split the interval \([-8,2]\) into two sub - intervals \([-8,-3]\) and \([-3,2]\). The net signed area \(A=\int_{-8}^{2}(3-\vert x + 3\vert)dx=\int_{-8}^{-3}(x + 6)dx+\int_{-3}^{2}(-x)dx\). First, calculate \(\int_{-8}^{-3}(x + 6)dx\). Using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)\), we have \(\int_{-8}^{-3}(x + 6)dx=\left[\frac{x^{2}}{2}+6x\right]_{-8}^{-3}\). \[ \begin{align*} \left(\frac{(-3)^{2}}{2}+6\times(-3)\right)-\left(\frac{(-8)^{2}}{2}+6\times(-8)\right)&=\left(\frac{9}{2}-18\right)-\left(32-48\right)\\ &=\frac{9}{2}-18 - 32 + 48\\ &=\frac{9}{2}-2\\ &=\frac{9 - 4}{2}=\frac{5}{2} \end{align*} \] Second, calculate \(\int_{-3}^{2}(-x)dx=\left[-\frac{x^{2}}{2}\right]_{-3}^{2}\). \[ \begin{align*} -\frac{2^{2}}{2}-\left(-\frac{(-3)^{2}}{2}\right)&=-2+\frac{9}{2}\\ &=\frac{-4 + 9}{2}=\frac{5}{2} \end{align*} \] ## Step3: Sum the results of the two integrals \(A=\frac{5}{2}+\frac{5}{2}=5\). # Answer: 5