find the net signed area between the graph of the function f(x)=|x - 1|-3 and the x - axis over the interval…

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})