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

Explanation:

Step1: Rewrite the absolute - value function

For (y = |x - 1|), we have (y=\begin{cases}x - 1, & x\geq1\1 - x, & x<1\end{cases}). So (f(x)=2-|x - 1|=\begin{cases}2-(x - 1)=3 - x, & x\geq1\2-(1 - x)=1 + x, & x<1\end{cases}).

Step2: Split the integral based on the break - point

We split the interval ([-6,4]) at (x = 1). The area (A=\int_{-6}^{1}(1 + x)dx+\int_{1}^{4}(3 - x)dx).

Step3: Calculate the first integral

Using the power rule (\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), (\int_{-6}^{1}(1 + x)dx=\left[x+\frac{x^{2}}{2}\right]{-6}^{1}=(1+\frac{1^{2}}{2})-(-6+\frac{(-6)^{2}}{2})=(1+\frac{1}{2})-(-6 + 18)=\frac{3}{2}-12=-\frac{21}{2}). But area is non - negative, so (\left|\int{-6}^{1}(1 + x)dx\right|=\left|\frac{3}{2}-12\right|=\frac{21}{2}).

Step4: Calculate the second integral

(\int_{1}^{4}(3 - x)dx=\left[3x-\frac{x^{2}}{2}\right]_{1}^{4}=(3\times4-\frac{4^{2}}{2})-(3\times1-\frac{1^{2}}{2})=(12 - 8)-(3-\frac{1}{2})=4-\frac{5}{2}=\frac{3}{2}).

Step5: Sum up the areas

(A=\frac{21}{2}+\frac{3}{2}=\frac{21 + 3}{2}=12).

Answer:

12