evaluate the following integral by interpreting it in terms of areas from drawing the graph of the function…

Explanation:

Step1: Analyze the function

The function (y = 3-\vert x\vert) can be written as (y=\begin{cases}3 - x, & x\geq0\3 + x, & x<0\end{cases}). It is a V - shaped graph with vertex at ((0,3)).

Step2: Split the integral by symmetry

Since (y = 3-\vert x\vert) is an even function ((f(-x)=f(x))), we can write (\int_{-4}^{4}(3 - \vert x\vert)dx = 2\int_{0}^{4}(3 - x)dx).

Step3: Use integral rules

We know that (\int_{0}^{4}(3 - x)dx=\int_{0}^{4}3dx-\int_{0}^{4}xdx). By the power - rule of integration (\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)) and (\int kdx=kx + C) ((k) is a constant), we have (\int_{0}^{4}3dx=3x\big|{0}^{4}=3\times4-3\times0 = 12) and (\int{0}^{4}xdx=\frac{x^{2}}{2}\big|{0}^{4}=\frac{4^{2}}{2}-\frac{0^{2}}{2}=8). So (\int{0}^{4}(3 - x)dx=12 - 8=4).

Step4: Calculate the original integral

Since (\int_{-4}^{4}(3 - \vert x\vert)dx = 2\int_{0}^{4}(3 - x)dx), then (\int_{-4}^{4}(3 - \vert x\vert)dx=2\times4 = 8).

Answer:

8