find the total area between the graph of the function f(x)=|x - 1|-2, graphed below, and the x - axis over…

Explanation:

Step1: Analyze the absolute - value function

The function (y = |x - 1|-2) can be written as a piece - wise function: (y=\begin{cases}x - 1-2=x - 3, &x\geq1\-(x - 1)-2=-x - 1, &x<1\end{cases}).

Step2: Find the x - intercepts

Set (y = 0). For (y=x - 3), (x=3) (when (x\geq1)); for (y=-x - 1), (x=-1) (when (x<1)).

Step3: Split the integral based on x - intercepts

We split the interval ([-4,6]) into sub - intervals ([-4,-1]), ([-1,3]), and ([3,6]) based on the x - intercepts (x=-1) and (x = 3). The area (A=\int_{-4}^{-1}(-(-x - 1))dx+\int_{-1}^{3}(-(|x - 1|-2))dx+\int_{3}^{6}(|x - 1|-2)dx). For (\int_{-4}^{-1}(-(-x - 1))dx=\int_{-4}^{-1}(x + 1)dx). Using the power rule (\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (\left[\frac{x^{2}}{2}+x\right]{-4}^{-1}=\left(\frac{(-1)^{2}}{2}-1\right)-\left(\frac{(-4)^{2}}{2}-4\right)=\left(\frac{1}{2}-1\right)-\left(8 - 4\right)=-\frac{1}{2}-4=-\frac{9}{2}), and the absolute value is (\frac{9}{2}). For (\int{-1}^{3}(-(|x - 1|-2))dx). When (x\in[-1,1]), (y=-(-x - 1 + 2)=x - 1); when (x\in[1,3]), (y=-(x - 1-2)=-x + 3). (\int_{-1}^{1}(x - 1)dx+\int_{1}^{3}(-x + 3)dx=\left[\frac{x^{2}}{2}-x\right]{-1}^{1}+\left[-\frac{x^{2}}{2}+3x\right]{1}^{3}=\left(\frac{1}{2}-1\right)-\left(\frac{1}{2}+1\right)+\left(-\frac{9}{2}+9\right)-\left(-\frac{1}{2}+3\right)=-2 + 2=2). For (\int_{3}^{6}(x - 3)dx=\left[\frac{x^{2}}{2}-3x\right]_{3}^{6}=\left(\frac{6^{2}}{2}-18\right)-\left(\frac{3^{2}}{2}-9\right)=0-\left(\frac{9}{2}-9\right)=\frac{9}{2}).

Step4: Sum up the areas of sub - regions

(A=\frac{9}{2}+2+\frac{9}{2}=11).

Answer:

11