find the net signed area between the graph of the function f(x)=3 - |x + 3| and the x - axis over the…

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