Mathematics › 🔬 Grade 12 › Riemann Sums: Building Area Out of Rectangles

Riemann Sums: Building Area Out of Rectangles

Cover the region under a curve with skinny rectangles, add them up, and let their number soar — the total converges to the exact area.

Grade 12Calculus / AP level

Riemann Sums: Building Area Out of Rectangles — illustration

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The big idea: There is no elementary formula for the area under a curved graph, so we approximate: slice the interval into strips and cap each with a rectangle whose height is the function somewhere on that strip. The sum of the rectangle areas is a Riemann sum. As the strips get thinner and more numerous, the approximation tightens onto a single value — the definite integral, the exact area.

🎯 By the end, you'll be able to

Approximate the area under a curve using left, right, and midpoint rectangles
Compute a Riemann sum for a given number of subintervals
Explain why left and right sums bracket the true area for a monotonic function
Describe the definite integral as the limit of Riemann sums as n → ∞

📎 You should already know

Area of a rectangle
Function evaluation and summation
The idea of a limit

The problem of curved area

Finding the area of a rectangle or triangle is easy. But what is the area under a curve — say beneath y = x² from x = 0 to x = 2, above the x-axis? No simple shape formula fits. The trick, going back to Archimedes, is to approximate the curved region with straight-sided pieces we can measure, then refine.

Slice into strips, cap with rectangles

Chop the interval [a, b] into n equal strips, each of width Δx = (b − a)⁄n. Over each strip, build a rectangle whose height is the function’s value at some point of that strip. Add up all the rectangle areas — height times Δx — and you have an estimate of the total area. This total is a Riemann sum.

🔑 A Riemann sum

A Riemann sum adds the areas of rectangles under a curve: each has width Δx and height f(xₕ) sampled at some point in the strip. More, thinner rectangles give a closer estimate of the true area.

\[ S_n = \sum_{i=1}^{n} f(x_i)\,\Delta x, \qquad \Delta x = \frac{b - a}{n} \]

The Riemann sum: total the rectangle areas, each height f(x_i) times the common width Delta-x.

🎮 Riemann Sums LIVE

Add more rectangles under a curve and watch the estimate converge to the true area.

Where do we sample the height?

The freedom is in where on each strip we read the height. A left sum uses the left edge of each strip, a right sum the right edge, and a midpoint sum the center. For a rising function, left edges sit below the curve (underestimate) and right edges above it (overestimate); the midpoint usually splits the difference and does better.

📝 Worked example: Estimate the area under y = x² on [0, 2] using 4 right-endpoint rectangles.

Width: \( \Delta x = (2 - 0)/4 = 0.5 \). Right endpoints: x = 0.5, 1, 1.5, 2.
Heights (square each): \( 0.25,\; 1,\; 2.25,\; 4 \).
Sum of heights: \( 0.25 + 1 + 2.25 + 4 = 7.5 \).
Multiply by the width: \( 7.5 \times 0.5 = 3.75 \).

✓ The right-sum estimate is <strong>3.75</strong> — an overestimate, since x² is increasing.

Refine, then take the limit

The true area under y = x² on [0, 2] is exactly 8⁄3 ≈ 2.667. Our left sum for n = 4 gives 1.75 (too low) and the right sum gives 3.75 (too high) — the truth is trapped between. Doubling n narrows the gap; pushing n to infinity closes it entirely. That limit is the definite integral.

\[ \int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x \]

The definite integral is the limit of Riemann sums as the rectangles become infinitely thin and numerous.

✨ Trapezoids and midpoints do better

Averaging the left and right sums gives the trapezoidal estimate — here (1.75 + 3.75)⁄2 = 2.75, already very near the true 2.667. The midpoint rule is typically even sharper. Smarter sampling converges faster, but every method shares the same limit: the exact integral.

📝 Worked example: Estimate the same area under y = x² on [0, 2] using 4 LEFT-endpoint rectangles.

Width \( \Delta x = 0.5 \). Left endpoints: x = 0, 0.5, 1, 1.5.
Heights: \( 0,\; 0.25,\; 1,\; 2.25 \).
Sum of heights: \( 0 + 0.25 + 1 + 2.25 = 3.5 \).
Multiply by the width: \( 3.5 \times 0.5 = 1.75 \).

✓ The left-sum estimate is <strong>1.75</strong> — an underestimate, bracketing the true area from below.

⚠️ For a rising curve: left under, right over

On an increasing function the left sum always underestimates and the right sum always overestimates the true area — the two bracket it. On a decreasing function it is the reverse. Knowing this lets you tell at a glance whether an estimate is high or low.

Check your understanding

1. As you use more and more rectangles, a Riemann sum estimate:

Thinner, more numerous rectangles hug the curve better, so the sum converges to the exact area — the definite integral.

2. For an increasing function, the RIGHT-endpoint sum is:

For a rising curve, the right edge of each strip is the tallest point, so the rectangles overshoot the true area.

3. For the interval [0, 2] with n = 4 rectangles, the width Δx is:

Δx = (b − a)/n = (2 − 0)/4 = 0.5.

4. The 4-rectangle LEFT sum for y = x² on [0, 2] is:

Left heights at x = 0, 0.5, 1, 1.5 are 0, 0.25, 1, 2.25; their sum 3.5 times Δx = 0.5 gives 1.75.

5. The definite integral of f from a to b is defined as:

The definite integral is the limit of the Riemann sums as the number of rectangles grows without bound.

✅ Key takeaways

Area under a curve has no simple formula, so we approximate it with rectangles.
A Riemann sum totals rectangle areas: each height f(xᵢ) times the width Δx = (b − a)/n.
Sampling heights at left, right, or midpoints gives left, right, and midpoint sums.
For a rising function the left sum underestimates and the right sum overestimates, bracketing the truth (e.g. 1.75 and 3.75 around 8/3 for x² on [0,2]).
Trapezoidal and midpoint rules converge faster, but all methods share the same limit.
As n → ∞ the rectangles become infinitely thin and the Riemann sum becomes the exact definite integral.