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

Explanation:

Step1: Analyze the absolute - value function

The function (y = 1-\vert x - 1\vert) can be written as a piece - wise function. When (x-1\geq0) (i.e., (x\geq1)), (y = 1-(x - 1)=2 - x); when (x - 1<0) (i.e., (x<1)), (y=1-(1 - x)=x).

Step2: Find the intersection points with the x - axis

Set (y = 0). For (y=x), (x = 0); for (y=2 - x), (x = 2).

Step3: Split the integral based on intervals

We split the interval ([-6,4]) into sub - intervals based on the behavior of the function and its intersection with the (x) - axis. We need to calculate the integral (\int_{-6}^{4}\vert1-\vert x - 1\vert\vert dx). We split it into three parts: (\int_{-6}^{0}-(1-\vert x - 1\vert)dx+\int_{0}^{2}(1-\vert x - 1\vert)dx+\int_{2}^{4}-(1-\vert x - 1\vert)dx). For (-6\leq x<0), (y=1-(1 - x)=x), and (\int_{-6}^{0}-x dx=-\left[\frac{x^{2}}{2}\right]{-6}^{0}=-\left(0-\frac{(-6)^{2}}{2}\right)=18). For (0\leq x<1), (y = x), and (\int{0}^{1}x dx=\left[\frac{x^{2}}{2}\right]{0}^{1}=\frac{1}{2}). For (1\leq x<2), (y=2 - x), and (\int{1}^{2}(2 - x)dx=\left[2x-\frac{x^{2}}{2}\right]{1}^{2}=(4 - 2)-(2-\frac{1}{2})=\frac{1}{2}). For (2\leq x<4), (y=-(2 - x)=x - 2), and (\int{2}^{4}(x - 2)dx=\left[\frac{x^{2}}{2}-2x\right]_{2}^{4}=(\frac{4^{2}}{2}-8)-(\frac{2^{2}}{2}-4)=2).

Step4: Sum up the areas

The total area (A=18+\frac{1}{2}+\frac{1}{2}+2 = 21).

Answer:

21