Evaluating a definite integral as the area of a triangle

The region under $f(x) = -2x+6$ from $x=0$ to $x=3$ is a right triangle. At $x=0$, $f(0)=6$, and at $x=3$, $f(3)=0$, so the triangle has vertices at $(0,0)$, $(0,6)$, and $(3,0)$.

The base of the triangle lies along the x-axis with length $3$, and the height along the y-axis is $6$. Since the region lies entirely above the x-axis, the integral equals the area: $\frac{1}{2}(3)(6) = 9$.

Therefore, $\int_0^3 (-2x+6)\,dx = 9$.

The height used was six. This came from evaluating f at x equals zero, since f of zero equals negative two times zero plus six, which is six. That value formed the vertical leg of the right triangle used to compute the area.