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

Explanation:

Step1: Rewrite absolute - value function

The absolute - value function (y = |x + 3|-3) can be rewritten as a piece - wise function. When (x+3\geq0) (i.e., (x\geq - 3)), (y=(x + 3)-3=x); when (x+3<0) (i.e., (x<-3)), (y=-(x + 3)-3=-x - 6).

Step2: Split the integral based on the break - point

We split the integral (\int_{-9}^{2}(|x + 3|-3)dx) into two integrals based on (x=-3). So (\int_{-9}^{2}(|x + 3|-3)dx=\int_{-9}^{-3}(-x - 6)dx+\int_{-3}^{2}x dx).

Step3: Integrate 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 - 6)dx=\left[-\frac{x^{2}}{2}-6x\right]_{-9}^{-3}). [ \begin{align*} &(-\frac{(-3)^{2}}{2}-6\times(-3))-(-\frac{(-9)^{2}}{2}-6\times(-9))\ =&(-\frac{9}{2}+18)-(-\frac{81}{2}+54)\ =&-\frac{9}{2}+18+\frac{81}{2}-54\ =&\frac{-9 + 81}{2}+18-54\ =&\frac{72}{2}+18-54\ =&36 + 18-54\ =&0 \end{align*} ]

Step4: Integrate the second integral

For (\int_{-3}^{2}x dx=\left[\frac{x^{2}}{2}\right]_{-3}^{2}=\frac{2^{2}}{2}-\frac{(-3)^{2}}{2}=\frac{4}{2}-\frac{9}{2}=-\frac{5}{2}).

Step5: Sum the results of the two integrals

(\int_{-9}^{2}(|x + 3|-3)dx=0-\frac{5}{2}=-\frac{5}{2}).

Answer:

(-\frac{5}{2})