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

Explanation:

Step1: Analyze the absolute - value function

First, rewrite (y = |x - 3|-2) 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 x - intercepts

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

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

We split the interval ([-6,6]) into sub - intervals ([-6,1]), ([1,3]), ([3,5]) and ([5,6]). The area (A=\int_{-6}^{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

For (\int_{-6}^{1}(x - 1)dx=\left[\frac{x^{2}}{2}-x\right]{-6}^{1}=\left(\frac{1^{2}}{2}-1\right)-\left(\frac{(-6)^{2}}{2}+6\right)=\frac{1}{2}-1-(18 + 6)=\frac{1 - 2}{2}-24=-\frac{1}{2}-24=-\frac{49}{2}), and the absolute value is (\frac{49}{2}). For (\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)=-\frac{3}{2}-\frac{1}{2}=-2), and the absolute value is (2). For (\int{3}^{5}(x - 5)dx=\left[\frac{x^{2}}{2}-5x\right]{3}^{5}=\left(\frac{5^{2}}{2}-25\right)-\left(\frac{3^{2}}{2}-15\right)=\frac{25}{2}-25-\frac{9}{2}+15=\frac{25 - 9}{2}-10=8 - 10=-2), and the absolute value is (2). For (\int{5}^{6}(x - 5)dx=\left[\frac{x^{2}}{2}-5x\right]_{5}^{6}=\left(\frac{6^{2}}{2}-30\right)-\left(\frac{5^{2}}{2}-25\right)=\left(18-30\right)-\left(\frac{25}{2}-25\right)=-12+\frac{25}{2}=\frac{1}{2}), and the absolute value is (\frac{1}{2}).

Step5: Sum up the areas of sub - intervals

(A=\frac{49}{2}+2 + 2+\frac{1}{2}=\frac{49 + 1}{2}+4=25+4 = 29).

Answer:

(29)