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

Explanation:

Step1: Rewrite the absolute - value function

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

Step2: Split the integral based on the break - point

We need to calculate (\int_{-4}^{4}(2-|x - 1|)dx=\int_{-4}^{1}(x + 1)dx+\int_{1}^{4}(3 - x)dx)

Step3: Calculate the first integral (\int_{-4}^{1}(x + 1)dx)

Using the power rule (\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (\int_{-4}^{1}(x + 1)dx=\left[\frac{x^{2}}{2}+x\right]_{-4}^{1}) [ \begin{align*} \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{3}{2}-4\ &=\frac{3 - 8}{2}=-\frac{5}{2} \end{align*} ] But we are looking for the area, so we take the absolute value (\left|-\frac{5}{2}\right|=\frac{5}{2})

Step4: Calculate the second integral (\int_{1}^{4}(3 - x)dx)

(\int_{1}^{4}(3 - x)dx=\left[3x-\frac{x^{2}}{2}\right]_{1}^{4}) [ \begin{align*} \left(3\times4-\frac{4^{2}}{2}\right)-\left(3\times1-\frac{1^{2}}{2}\right)&=(12 - 8)-\left(3-\frac{1}{2}\right)\ &=4-\frac{5}{2}\ &=\frac{8 - 5}{2}=\frac{3}{2} \end{align*} ] Take the absolute value (\left|\frac{3}{2}\right|=\frac{3}{2})

Step5: Calculate the total area

The total area (A=\int_{-4}^{1}|x + 1|dx+\int_{1}^{4}|3 - x|dx) We also can consider the geometric approach. The function (y = 2-|x - 1|) is a V - shaped function. The vertex of the V is at ((1,2)). The left - hand side of the V for (x<1) has a slope of 1 and the right - hand side for (x\geq1) has a slope of - 1. We can split the interval ([-4,4]) into two parts. The area of the left - hand triangle (from (x=-4) to (x = 1)): The base of the triangle is (1-(-4)=5) and the height is (y(-4)=-4 + 1=-3) (but we take the absolute value of the height), and the area of a triangle (A_1=\frac{1}{2}\times5\times3=\frac{15}{2}) The area of the right - hand triangle (from (x = 1) to (x = 4)): The base of the triangle is (4 - 1=3) and the height is (y(4)=3-4=-1) (take the absolute value), and the area of a triangle (A_2=\frac{1}{2}\times3\times1=\frac{3}{2}) The total area (A=\frac{15}{2}+\frac{3}{2}=18)

Answer:

18