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

Explanation:

Step1: Rewrite the absolute - value function

For (y = |x - 2|-1), when (x\geq2), (y=(x - 2)-1=x - 3); when (x<2), (y=-(x - 2)-1=-x + 1).

Step2: Split the integral based on the break - point

The break - point of (y = |x - 2|-1) is (x = 2). We split the integral (\int_{-2}^{6}| |x - 2|-1|dx) into three parts: (\int_{-2}^{1}(-(-x + 1))dx+\int_{1}^{2}(-x + 1)dx+\int_{2}^{5}(x - 3)dx+\int_{5}^{6}(x - 3)dx). First, for (\int_{-2}^{1}(-(-x + 1))dx=\int_{-2}^{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]{-2}^{1}=\left(\frac{1}{2}-1\right)-\left(\frac{4}{2}+2\right)=\frac{1 - 2}{2}-(2 + 2)=-\frac{1}{2}-4=-\frac{9}{2}), and its absolute value is (\frac{9}{2}). Second, (\int{1}^{2}(-x + 1)dx=\left[-\frac{x^{2}}{2}+x\right]{1}^{2}=\left(-\frac{4}{2}+2\right)-\left(-\frac{1}{2}+1\right)=0-\frac{1}{2}=-\frac{1}{2}), and its absolute value is (\frac{1}{2}). Third, (\int{2}^{5}(x - 3)dx=\left[\frac{x^{2}}{2}-3x\right]{2}^{5}=\left(\frac{25}{2}-15\right)-\left(\frac{4}{2}-6\right)=\frac{25 - 30}{2}-\left(2 - 6\right)=-\frac{5}{2}+4=\frac{3}{2}), and its absolute value is (\frac{3}{2}). Fourth, (\int{5}^{6}(x - 3)dx=\left[\frac{x^{2}}{2}-3x\right]_{5}^{6}=\left(\frac{36}{2}-18\right)-\left(\frac{25}{2}-15\right)=0-\left(\frac{25 - 30}{2}\right)=\frac{5}{2}), and its absolute value is (\frac{5}{2}).

Step3: Sum up the absolute - value of the integrals

(A=\frac{9}{2}+\frac{1}{2}+\frac{3}{2}+\frac{5}{2}=\frac{9 + 1+3 + 5}{2}=9).

Answer:

9