Question

evaluate the following integral by interpreting it in terms of areas from drawing the graph of the function being integrated. $int_{-4}^{4}(3 - |x|)dx=$

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 \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\) 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