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

Explanation:

Step1: Analyze the absolute - value function

First, rewrite the absolute - value function $y = 1-\vert x - 1\vert$ 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: Split the integral based on the break - point

The break - point of the piece - wise function is $x = 1$. We split the integral over the interval $[-3,5]$ into three parts: $[-3,1]$, $[1,2]$, and $[2,5]$. The area $A=\int_{-3}^{1}x dx+\int_{1}^{2}(2 - x)dx-\int_{2}^{5}(2 - x)dx$.

Step3: Calculate the first integral

Use the power rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$. $\int_{-3}^{1}x dx=\left[\frac{x^{2}}{2}\right]{-3}^{1}=\frac{1^{2}}{2}-\frac{(-3)^{2}}{2}=\frac{1}{2}-\frac{9}{2}=-4$. But we are interested in the area, so we take the absolute value $\left|\int{-3}^{1}x dx\right| = 4$.

Step4: Calculate the second integral

$\int_{1}^{2}(2 - x)dx=\int_{1}^{2}2dx-\int_{1}^{2}x dx$. $\int_{1}^{2}2dx=2x\big|{1}^{2}=2\times(2 - 1)=2$, $\int{1}^{2}x dx=\left[\frac{x^{2}}{2}\right]{1}^{2}=\frac{2^{2}}{2}-\frac{1^{2}}{2}=\frac{4 - 1}{2}=\frac{3}{2}$. So, $\int{1}^{2}(2 - x)dx=2-\frac{3}{2}=\frac{1}{2}$.

Step5: Calculate the third integral

$\int_{2}^{5}(2 - x)dx=\int_{2}^{5}2dx-\int_{2}^{5}x dx$. $\int_{2}^{5}2dx=2x\big|{2}^{5}=2\times(5 - 2)=6$, $\int{2}^{5}x dx=\left[\frac{x^{2}}{2}\right]{2}^{5}=\frac{5^{2}}{2}-\frac{2^{2}}{2}=\frac{25 - 4}{2}=\frac{21}{2}$. So, $\int{2}^{5}(2 - x)dx=6-\frac{21}{2}=-\frac{9}{2}$. Take the absolute value $\left|\int_{2}^{5}(2 - x)dx\right|=\frac{9}{2}$.

Step6: Sum up the areas

$A = 4+\frac{1}{2}+\frac{9}{2}=4 + 5=9$.

Answer:

$9$