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

The absolute - value function (y = |x - 3|-4) can be rewritten as a piece - wise function. When (x-3\geq0) (i.e., (x\geq3)), (y=(x - 3)-4=x - 7); when (x - 3<0) (i.e., (x<3)), (y=-(x - 3)-4=-x - 1).

Step2: Split the integral based on the break - point

We want to find the net signed area (A=\int_{-3}^{9}(|x - 3|-4)dx=\int_{-3}^{3}(-x - 1)dx+\int_{3}^{9}(x - 7)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_{-3}^{3}(-x - 1)dx=\left[-\frac{x^{2}}{2}-x\right]_{-3}^{3}). [ \begin{align*} \left(-\frac{3^{2}}{2}-3\right)-\left(-\frac{(-3)^{2}}{2}-(-3)\right)&=\left(-\frac{9}{2}-3\right)-\left(-\frac{9}{2}+3\right)\ &=-\frac{9 + 6}{2}-\left(-\frac{9 - 6}{2}\right)\ &=-\frac{15}{2}+\frac{3}{2}\ &=-6 \end{align*} ]

Step4: Integrate the second integral

For (\int_{3}^{9}(x - 7)dx=\left[\frac{x^{2}}{2}-7x\right]_{3}^{9}). [ \begin{align*} \left(\frac{9^{2}}{2}-7\times9\right)-\left(\frac{3^{2}}{2}-7\times3\right)&=\left(\frac{81}{2}-63\right)-\left(\frac{9}{2}-21\right)\ &=\frac{81-126}{2}-\frac{9 - 42}{2}\ &=-\frac{45}{2}+\frac{33}{2}\ &=-6 \end{align*} ]

Step5: Calculate the net signed area

(A=-6+( - 6)=-12).

Answer:

(-12)