Evaluating a definite integral via the area of a triangle

The graph of $f(x) = 3-|x-2|$ crosses the x-axis where $3 - |x-2| = 0$, i.e., $|x-2| = 3$, giving $x = -1$ or $x = 5$. However, we only need the portion on $[0,4]$: at $x=0$, $f(0) = 3-2=1$, and at $x=4$, $f(4) = 3-2=1$, while the peak occurs at $x=2$ with $f(2) = 3$. So the region on $[0,4]$ is a triangle-like shape, but since $f$ stays positive throughout (never touching zero on this interval), it's actually a trapezoid... Let's instead directly compute using the fact that the shape is a triangle centered at $x=2$: the full triangle (base from $x=-1$ to $x=5$, height 3) is symmetric, and $[0,4]$ captures the full triangle minus two small corner triangles.

Simpler approach: split the region at the peak $x=2$. On $[0,2]$, $f(x) = 3-(2-x) = x+1$, a line from $(0,1)$ to $(2,3)$ — this is a trapezoid with parallel sides $1$ and $3$ and width $2$, area $= \frac{1}{2}(1+3)(2) = 4$. On $[2,4]$, $f(x) = 3-(x-2) = 5-x$, a line from $(2,3)$ to $(4,1)$ — again a trapezoid with parallel sides $3$ and $1$ and width $2$, area $=4$. Since $f(x) \geq 0$ on all of $[0,4]$, the integral equals the total area: $4+4 = 8$.