question find the total area between the grap...

# 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