question find the net signed area between the...
question find the net signed area between the graph of the function f(x)=|x - 3|-4 and the x - axis over the interval -3,9, illustrated in the following image. submit your answer as an exact value.
Answer
# 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:
\(-12\)