Left Riemann sum setup and value

With $a=1$, $b=5$, $n=4$, the width is $\Delta x = \frac{5-1}{4} = 1$. The subintervals are $[1,2],[2,3],[3,4],[4,5]$, so the left endpoints are $x_0=1, x_1=2, x_2=3, x_3=4$.

The left sum is $L_4 = \Delta x \left[f(1)+f(2)+f(3)+f(4)\right] = 1\left[1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right] = \frac{25}{12} \approx 2.083$.

The right endpoints of the subintervals $[1,2],[2,3],[3,4],[4,5]$ are $x_1=2, x_2=3, x_3=4, x_4=5$, with $\Delta x = 1$.

The right sum is $R_4 = \Delta x\left[f(2)+f(3)+f(4)+f(5)\right] = 1\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right] = \frac{77}{60} \approx 1.283$.

Since $f$ is decreasing on $[1,5]$, on each subinterval the left endpoint gives the largest function value and the right endpoint gives the smallest function value.

Therefore the left-endpoint rectangles always lie above the curve (overestimating the area), while the right-endpoint rectangles lie below the curve (underestimating the area). So $L_4$ is an overestimate and $R_4$ is an underestimate, consistent with $L_4 \approx 2.083 > R_4 \approx 1.283$.

The left sum is bigger than the right sum. Since f of x equals one over x is decreasing on the interval, left endpoints give larger function values than right endpoints on every subinterval. This makes the left-endpoint sum an overestimate and the right-endpoint sum an underestimate. This matches the earlier computed values of about two point zero eight three versus about one point two eight three.