Signed area for a triangle above the x-axis

On $[0,4]$, the graph of $f(x)=4-x$ starts at the point $(0,4)$ and decreases linearly to $(4,0)$, forming a right triangle with the x-axis. The base of this triangle runs from $x=0$ to $x=4$, so the base length is $4$, and the height is $f(0) = 4$.

Since $f(x) \ge 0$ on this interval, the region lies entirely above the x-axis, so the integral equals the area of the triangle: $\frac{1}{2}(4)(4) = 8$. Thus $\int_0^4 (4-x)\,dx = 8$.

On $[4,6]$, the graph goes from $(4,0)$ to $(6,-2)$, forming a right triangle below the x-axis with base $2$ (from $x=4$ to $x=6$) and height $2$ (the vertical drop to $f(6)=-2$).

The area of this triangle is $\frac{1}{2}(2)(2) = 2$, but since the region lies below the x-axis, the signed area (and hence the integral) is negative: $\int_4^6 (4-x)\,dx = -2$.