Left sum over- or under-estimate on [0,4]

First, note that $f(x) = 4x - x^2$ is a downward parabola with vertex at $x = 2$, where $f(2) = 4$. On $[0,2]$, $f$ is increasing, and on $[2,4]$, $f$ is decreasing, so the behavior of $f$ changes at $x=2$.

On the increasing portion $[0,2]$, the left endpoint of each subinterval gives a smaller function value than points to its right, so the left-endpoint rectangles lie below the curve, underestimating area there. On the decreasing portion $[2,4]$, the left endpoint gives a larger function value than points to its right, so the rectangles lie above the curve, overestimating area there.

To determine the overall effect, note that on $[0,2]$, there are 4 subintervals of width 0.5 (from $x=0$ to $x=2$), and on $[2,4]$, there are also 4 subintervals of width 0.5. The function $f(x) = 4x - x^2$ is symmetric about the line $x=2$ in the sense that $f(2-t) = f(2+t)$ for all $t$. This symmetry means that the underestimation error on the left half mirrors the overestimation error on the right half in magnitude. Since the two halves have equal width and the errors are equal in magnitude but opposite in sign, they cancel exactly. Therefore, the left-endpoint Riemann sum neither overestimates nor underestimates the true area; the two types of errors balance out.