Riemann Sum — Midpoint Approximation

The midpoint Riemann sum uses the function value at the center of each subinterval. For most functions this is dramatically more accurate than left or right sums — the over-and-under errors of the rectangle's two halves tend to cancel.

The Setup

The midpoint of the $i$-th subinterval is:

$x_i^{*} = a + \left(i + \tfrac{1}{2}\right)\Delta x$

The midpoint Riemann sum is:

$M_n = \sum_{i=0}^{n-1} f(x_i^{*})\,\Delta x$

Step-by-Step Solution

Given: $f(x) = x^{2}$, $a = 0$, $b = 4$, $n = 8$.

Find: The midpoint Riemann sum $M_8$.

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Step 1 — Compute $\Delta x$.

$\Delta x = \dfrac{4 - 0}{8} = 0.5$

Step 2 — List the midpoints.

The first midpoint is at $0 + (0 + 0.5)\cdot 0.5 = 0.25$, then they spread out by $\Delta x = 0.5$:

$x_i^{*} = 0.25,\; 0.75,\; 1.25,\; 1.75,\; 2.25,\; 2.75,\; 3.25,\; 3.75$

Step 3 — Evaluate $f(x_i^{*}) = (x_i^{*})^{2}$.

$(0.25)^{2} = 0.0625, \;\;\; (0.75)^{2} = 0.5625, \;\;\; (1.25)^{2} = 1.5625, \;\;\; (1.75)^{2} = 3.0625,$

$(2.25)^{2} = 5.0625, \;\;\; (2.75)^{2} = 7.5625, \;\;\; (3.25)^{2} = 10.5625, \;\;\; (3.75)^{2} = 14.0625$

Step 4 — Sum the function values.

$\sum_{i=0}^{7} f(x_i^{*}) = 0.0625 + 0.5625 + 1.5625 + 3.0625 + 5.0625 + 7.5625 + 10.5625 + 14.0625 = 42.5$

Step 5 — Multiply by $\Delta x$.

$M_8 = 0.5 \times 42.5 = 21.25$

Step 6 — Compare to the exact value.

$\int_{0}^{4} x^{2}\,dx = \dfrac{64}{3} \approx 21.333$

The error is only $21.333 - 21.25 = 0.083$ — about 0.4%! Compare this to the left sum's 18% error and the right sum's 20% error with the exact same number of subdivisions. The midpoint rule is dramatically better.

Why? Inside each subinterval, the curve is approximately linear. The midpoint rectangle's height matches the line's average value over the interval — so it counts the high half of the interval and the low half exactly equally. The error is proportional to $f''$ (concavity) rather than $f'$ (slope), which makes it converge much faster as $n \to \infty$.

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Answer: The midpoint Riemann sum with $n = 8$ rectangles is

$\boxed{\;M_8 = 21.25\;}$

This is within 0.4% of the exact value $\dfrac{64}{3} \approx 21.333$. With the same number of rectangles, midpoint dominates left and right sums. Even better is Simpson's rule, which combines midpoint and trapezoidal sums to get error proportional to $f^{(4)}$.

Try It

• Adjust n — see how quickly the midpoint error vanishes (it shrinks like $1/n^{2}$, twice as fast as left/right).
• Toggle show exact area to compare against the true region.
• Notice how each rectangle's top spans both above and below the curve — the over-counting on one side cancels the under-counting on the other.