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

# Explanation: ## Step1: Find where the function crosses x - axis Set $f(x)=-x - 1=0$, then $x=-1$. ## Step2: Split the integral based on the x - intercept The area $A=\int_{-6}^{-1}(-x - 1)dx+\left|\int_{-1}^{3}(-x - 1)dx\right|$. ## Step3: Calculate $\int_{-6}^{-1}(-x - 1)dx$ Using the power - rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(-x - 1)dx=-\frac{x^{2}}{2}-x+C$. Evaluating $\left(-\frac{x^{2}}{2}-x\right)\big|_{-6}^{-1}=\left(-\frac{(-1)^{2}}{2}-(-1)\right)-\left(-\frac{(-6)^{2}}{2}-(-6)\right)=\left(-\frac{1}{2}+1\right)-\left(-18 + 6\right)=\frac{1}{2}+12=\frac{25}{2}$. ## Step4: Calculate $\int_{-1}^{3}(-x - 1)dx$ Evaluating $\left(-\frac{x^{2}}{2}-x\right)\big|_{-1}^{3}=\left(-\frac{3^{2}}{2}-3\right)-\left(-\frac{(-1)^{2}}{2}-(-1)\right)=\left(-\frac{9}{2}-3\right)-\left(-\frac{1}{2}+1\right)=-\frac{15}{2}-\frac{1}{2}=-8$. Then $\left|\int_{-1}^{3}(-x - 1)dx\right| = 8$. ## Step5: Calculate the total area $A=\frac{25}{2}+8=\frac{25 + 16}{2}=\frac{41}{2}$. # Answer: $\frac{41}{2}$