To evaluate the integral [tex]\(\int_0^6 f(x) \, dx\)[/tex], where the function [tex]\( f(x) \)[/tex] is given by:
[tex]\[
f(x) =
\begin{cases}
4 - x & \text{if } x \leq 4 \\
2x - 8 & \text{if } x > 4
\end{cases},
\][/tex]
we can split the integral at the point where the definition of [tex]\( f(x) \)[/tex] changes, which is [tex]\( x = 4 \)[/tex]. So we break the integral into two parts:
[tex]\[
\int_0^6 f(x) \, dx = \int_0^4 (4 - x) \, dx + \int_4^6 (2x - 8) \, dx
\][/tex]
First, evaluate the integral [tex]\(\int_0^4 (4 - x) \, dx\)[/tex]:
The graph of [tex]\( 4 - x \)[/tex] from [tex]\( x = 0 \)[/tex] to [tex]\( x = 4 \)[/tex] forms a right triangle with a base of 4 (along the x-axis) and a height of 4 (along the y-axis). The area of a triangle is given by [tex]\(\frac{1}{2} \times \text{base} \times \text{height}\)[/tex]:
[tex]\[
\text{Area}_1 = \frac{1}{2} \times 4 \times 4 = 8.0
\][/tex]
Next, evaluate the integral [tex]\(\int_4^6 (2x - 8) \, dx\)[/tex]:
The graph of [tex]\( 2x - 8 \)[/tex] from [tex]\( x = 4 \)[/tex] to [tex]\( x = 6 \)[/tex] forms another right triangle. At [tex]\( x = 4 \)[/tex], [tex]\( f(x) = 0 \)[/tex], and at [tex]\( x = 6 \)[/tex], [tex]\( f(x) = 2(6) - 8 = 4 \)[/tex]. The base of this triangle is [tex]\( 6 - 4 = 2 \)[/tex] and the height is [tex]\( 4 \)[/tex]. Using the area formula for a triangle:
[tex]\[
\text{Area}_2 = \frac{1}{2} \times 2 \times 4 = 4.0
\][/tex]
Finally, sum the areas of the two triangles to find the total area under the curve from [tex]\( x = 0 \)[/tex] to [tex]\( x = 6 \)[/tex]:
[tex]\[
\text{Total Area} = \text{Area}_1 + \text{Area}_2 = 8.0 + 4.0 = 12.0
\][/tex]
Therefore, the value of the integral [tex]\(\int_0^6 f(x) \, dx\)[/tex] is:
[tex]\[
\boxed{12}
\][/tex]
