question find the net signed area between the graph of the function f(x)=|x + 3|-4 and the x - axis over the…

Explanation:

Step1: Rewrite the absolute - value function

For (y = |x + 3|-4), when (x+3\geq0) (i.e., (x\geq - 3)), (y=(x + 3)-4=x - 1); when (x+3<0) (i.e., (x<-3)), (y=-(x + 3)-4=-x - 7).

Step2: Split the integral based on the break - point

We split the integral (\int_{-9}^{3}(|x + 3|-4)dx=\int_{-9}^{-3}(-x - 7)dx+\int_{-3}^{3}(x - 1)dx).

Step3: Calculate the first integral

Using the power rule (\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), for (\int_{-9}^{-3}(-x - 7)dx=\left[-\frac{x^{2}}{2}-7x\right]_{-9}^{-3}). [ \begin{align*} &(-\frac{(-3)^{2}}{2}-7\times(-3))-(-\frac{(-9)^{2}}{2}-7\times(-9))\ =&(-\frac{9}{2}+21)-(-\frac{81}{2}+63)\ =&(-\frac{9}{2}+21+\frac{81}{2}-63)\ =&\frac{-9 + 81}{2}+21-63\ =&\frac{72}{2}+21 - 63\ =&36+21-63\ =& - 6 \end{align*} ]

Step4: Calculate the second integral

For (\int_{-3}^{3}(x - 1)dx=\left[\frac{x^{2}}{2}-x\right]_{-3}^{3}). [ \begin{align*} &(\frac{3^{2}}{2}-3)-(\frac{(-3)^{2}}{2}-(-3))\ =&(\frac{9}{2}-3)-(\frac{9}{2}+3)\ =&\frac{9}{2}-3-\frac{9}{2}-3\ =& - 6 \end{align*} ]

Step5: Find the net - signed area

Add the results of the two integrals: (\int_{-9}^{3}(|x + 3|-4)dx=-6+( - 6)=-12).

Answer:

(-12)