find the net signed area between the graph of...
find the net signed area between the graph of the function f(x)=|x - 1|-3 and the x - axis over the interval -5,6. submit your answer as an exact value. provide your answer below: net signed area
Answer
# Explanation:
## Step1: Rewrite the absolute - value function
The absolute - value function \(y = |x - 1|-3\) can be rewritten as a piece - wise function.
When \(x-1\geq0\) (i.e., \(x\geq1\)), \(y=(x - 1)-3=x - 4\).
When \(x - 1<0\) (i.e., \(x<1\)), \(y=-(x - 1)-3=-x - 2\).
## Step2: Split the integral based on the break - point
We need to calculate \(\int_{-5}^{6}(|x - 1|-3)dx=\int_{-5}^{1}(-x - 2)dx+\int_{1}^{6}(x - 4)dx\).
## Step3: Calculate the first integral \(\int_{-5}^{1}(-x - 2)dx\)
Using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)\), we have:
\(\int_{-5}^{1}(-x - 2)dx=-\int_{-5}^{1}x dx-2\int_{-5}^{1}dx\)
\(=-\left[\frac{x^{2}}{2}\right]_{-5}^{1}-2[x]_{-5}^{1}\)
\(=-\left(\frac{1^{2}}{2}-\frac{(-5)^{2}}{2}\right)-2(1-( - 5))\)
\(=-\left(\frac{1}{2}-\frac{25}{2}\right)-2\times6\)
\(=-\left(\frac{1 - 25}{2}\right)-12\)
\(=-\left(-\frac{24}{2}\right)-12\)
\(=12 - 12=0\).
## Step4: Calculate the second integral \(\int_{1}^{6}(x - 4)dx\)
\(\int_{1}^{6}(x - 4)dx=\int_{1}^{6}x dx-4\int_{1}^{6}dx\)
\(=\left[\frac{x^{2}}{2}\right]_{1}^{6}-4[x]_{1}^{6}\)
\(=\frac{6^{2}}{2}-\frac{1^{2}}{2}-4(6 - 1)\)
\(=\frac{36}{2}-\frac{1}{2}-20\)
\(=\frac{36 - 1}{2}-20\)
\(=\frac{35}{2}-20\)
\(=\frac{35 - 40}{2}=-\frac{5}{2}\).
## Step5: Find the net - signed area
\(\int_{-5}^{6}(|x - 1|-3)dx=0-\frac{5}{2}=-\frac{5}{2}\).
# Answer:
\(-\frac{5}{2}\)