Riemann Sum Variants, and Which Way They Err
- Compute left, right, midpoint, and trapezoidal approximations
- Determine whether an approximation over- or under-estimates from monotonicity or concavity
- Handle unequal subintervals, as tables usually provide
Four approximations, one idea
Each variant approximates the area under a curve with rectangles or trapezoids, differing only in which height is used. A left Riemann sum uses the function value at the left edge of each subinterval; a right sum uses the right edge; a midpoint sum uses the value at the center. A trapezoidal sum uses the average of the two endpoint values on each subinterval, which is equivalent to averaging the left and right sums when the subintervals are equal. In every case the contribution of one subinterval is height × width, and the total is the sum. Nothing about this requires equal widths, which is important because tables rarely supply them.
Over or under: two separate arguments
The exam asks whether an approximation over- or under-estimates the true integral, and there are two distinct arguments — use the one that fits the variant. For left and right sums, the deciding property is monotonicity. If f is increasing, the left endpoint is the lowest value on each subinterval, so the left sum underestimates and the right sum overestimates; if f is decreasing, the reverse. For the trapezoidal sum, the deciding property is concavity. If f is concave up, each chord lies above the curve, so the trapezoidal sum overestimates; if concave down, it underestimates. The midpoint rule runs opposite to the trapezoid on a concave-up function — it underestimates — because the tangent-like rectangle at the midpoint cuts below the chord. Naming the property you are using is part of the justification.
Unequal subintervals are the normal case
Table-based questions almost always give unequal spacing, and treating the widths as equal is the standard error. Compute each subinterval's width separately and multiply by the height that variant calls for. For a trapezoidal approximation with unequal widths, the contribution of a subinterval from x₀ to x₁ is [(f(x₀) + f(x₁))/2]·(x₁ − x₀) — average the heights, then multiply by that particular width. Do not reach for the (h/2)[f₀ + 2f₁ + 2f₂ + … + fₙ] shortcut, which is valid only for equal widths and will silently give a wrong answer otherwise. Note also that a midpoint sum from a table is only possible when the table happens to contain the midpoints.
A table gives f(0) = 3, f(2) = 5, f(5) = 8, f(6) = 12. Use a trapezoidal approximation with the three subintervals to estimate the integral of f from 0 to 6, and state whether it over- or under-estimates if f is concave up.
- 1.Subinterval [0, 2] has width 2 and heights 3 and 5: [(3 + 5)/2]·2 = 4·2 = 8.
- 2.Subinterval [2, 5] has width 3 and heights 5 and 8: [(5 + 8)/2]·3 = 6.5·3 = 19.5.
- 3.Subinterval [5, 6] has width 1 and heights 8 and 12: [(8 + 12)/2]·1 = 10·1 = 10.
- 4.Sum the three contributions: 8 + 19.5 + 10 = 37.5.
- 5.If f is concave up, every chord lies above the curve, so each trapezoid contains more area than the true region — the approximation overestimates.
Widths of 2, 3 and 1 in that example are not a trick — unequal spacing is what real tables look like. Check the widths before computing anything, and never assume the h/2 shortcut applies.
If f is increasing and concave down on [a, b], a left Riemann sum will
For a function that is concave up on [a, b], the trapezoidal approximation
Answer the 2 checkpoints as you read.
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