find the total area between the graph of the function f(x)=|x - 3|-2 graphed below, and the x - axis over…

Explanation:

Step1: Rewrite the absolute - value function

The function (y = |x - 3|-2) can be rewritten as a piece - wise function. When (x-3\geq0) (i.e., (x\geq3)), (y=(x - 3)-2=x - 5); when (x - 3<0) (i.e., (x<3)), (y=-(x - 3)-2=-x + 1).

Step2: Find the intersection points with the x - axis

Set (y = 0). For (y=-x + 1), when (y = 0), (x = 1). For (y=x - 5), when (y = 0), (x = 5).

Step3: Split the integral based on the intervals

We need to split the integral (\int_{-4}^{6}| |x - 3|-2|dx) into sub - integrals based on the intervals ([-4,1]), ([1,3]), ([3,5]) and ([5,6]). On ([-4,1]), (y=-x + 1\geq0), the area (A_1=\int_{-4}^{1}(-x + 1)dx=\left[-\frac{x^{2}}{2}+x\right]{-4}^{1}=-\frac{1}{2}+1-\left(-\frac{(-4)^{2}}{2}-4\right)=-\frac{1}{2}+1 + 8 + 4=\frac{-1 + 2+16 + 8}{2}=\frac{25}{2}). On ([1,3]), (y=-x + 1\leq0), the area (A_2=\int{1}^{3}(x - 1)dx=\left[\frac{x^{2}}{2}-x\right]{1}^{3}=\frac{9}{2}-3-\left(\frac{1}{2}-1\right)=\frac{9 - 6-1 + 2}{2}=2). On ([3,5]), (y=x - 5\leq0), the area (A_3=\int{3}^{5}(5 - x)dx=\left[5x-\frac{x^{2}}{2}\right]{3}^{5}=25-\frac{25}{2}-\left(15-\frac{9}{2}\right)=25-\frac{25}{2}-15+\frac{9}{2}=\frac{50 - 25-30 + 9}{2}=2). On ([5,6]), (y=x - 5\geq0), the area (A_4=\int{5}^{6}(x - 5)dx=\left[\frac{x^{2}}{2}-5x\right]_{5}^{6}=\frac{36}{2}-30-\left(\frac{25}{2}-25\right)=18-30-\frac{25}{2}+25=\frac{36 - 60-25 + 50}{2}=\frac{1}{2}).

Step4: Calculate the total area

The total area (A=A_1+A_2+A_3+A_4=\frac{25}{2}+2 + 2+\frac{1}{2}=\frac{25 + 4+4 + 1}{2}=17).

Answer:

17