find the total area between the graph of the ...

# 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