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

Explanation:

Step1: Analyze the absolute - value function

Rewrite (y = |x - 3|-2) as a piece - wise function. When (x\geq3), (y=(x - 3)-2=x - 5); when (x<3), (y=-(x - 3)-2=-x + 1).

Step2: Find the x - intercepts

For (y=-x + 1), set (y = 0), then (0=-x + 1), (x = 1). For (y=x - 5), set (y = 0), then (0=x - 5), (x = 5).

Step3: Split the integral based on x - intercepts and intervals

We need to split the interval ([-4,6]) into sub - intervals based on the x - intercepts and the behavior of the function. We split it into ([-4,1]), ([1,3]), ([3,5]) and ([5,6]). The area (A=\int_{-4}^{1}(-(-x + 1))dx+\int_{1}^{3}(-x + 1)dx+\int_{3}^{5}(-(x - 5))dx+\int_{5}^{6}(x - 5)dx).

Step4: Calculate each integral

Integral 1: (\int_{-4}^{1}(x - 1)dx=\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}{2}-12=-\frac{1}{2}-12=-\frac{25}{2}). But we take the absolute - value, so (A_1=\frac{25}{2}).

Integral 2: (\int_{1}^{3}(-x + 1)dx=\left[-\frac{x^{2}}{2}+x\right]_{1}^{3}=\left(-\frac{3^{2}}{2}+3\right)-\left(-\frac{1^{2}}{2}+1\right)=\left(-\frac{9}{2}+3\right)-\left(-\frac{1}{2}+1\right)=\left(-\frac{9 - 6}{2}\right)-\left(\frac{-1 + 2}{2}\right)=-\frac{3}{2}-\frac{1}{2}=-2). Take the absolute - value, (A_2 = 2).

Integral 3: (\int_{3}^{5}(-(x - 5))dx=\int_{3}^{5}(-x + 5)dx=\left[-\frac{x^{2}}{2}+5x\right]_{3}^{5}=\left(-\frac{5^{2}}{2}+5\times5\right)-\left(-\frac{3^{2}}{2}+5\times3\right)=\left(-\frac{25}{2}+25\right)-\left(-\frac{9}{2}+15\right)=\frac{-25 + 50}{2}-\frac{-9 + 30}{2}=\frac{25}{2}-\frac{21}{2}=2).

Integral 4: (\int_{5}^{6}(x - 5)dx=\left[\frac{x^{2}}{2}-5x\right]_{5}^{6}=\left(\frac{6^{2}}{2}-5\times6\right)-\left(\frac{5^{2}}{2}-5\times5\right)=\left(18-30\right)-\left(\frac{25}{2}-25\right)=-12-\left(\frac{25 - 50}{2}\right)=-12+\frac{25}{2}=\frac{-24 + 25}{2}=\frac{1}{2}).

Step5: Sum up the areas

(A=\frac{25}{2}+2 + 2+\frac{1}{2}=\frac{25 + 4+4 + 1}{2}=\frac{34}{2}=17).

Answer:

17