Module 5 · Lesson 5.2

The Definite Integral & Riemann Sums

Discover how slicing the area under a curve into thin rectangles — and letting them get infinitely thin — defines the definite integral.

Intuition

If a curve were a straight edge, finding the area underneath it would just be geometry: a rectangle, a triangle, done. But most curves bend, so there's no single clean shape to measure. The fix is to chop the region into thin vertical strips, treat each strip as a rectangle you already know how to measure, and add them all up.

Core concept

Each rectangle's area is $f(x_i)\cdot\Delta x$ — height times width. Adding up $n$ of these rectangles gives an estimate of the area under the curve. The estimate isn't exact, because rectangles don't perfectly trace a curved edge. But as $n$ grows and each rectangle gets thinner, the leftover gaps shrink toward zero, and the sum approaches the true area exactly.

f(x) = x, four rectangles
Function & intervalf(x) = x, from 0 to 4
Rectangle width Δx1
Right-edge heightsf(1), f(2), f(3), f(4)
Sum of areas1 + 2 + 3 + 4 = 10
Exact area (triangle)½ × 4 × 4 = 8
AnswerEstimate overshoots by 2

Interactive Lab

Slide to change how many rectangles slice the region under the curve. Watch the gaps between the bars and the curve shrink, and see the running sum close in on the exact area.

x y
n = 4 Riemann sum ≈ — Exact area = 9.33

Key Idea

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

The definite integral is exactly the limit of the Riemann sum as the rectangles become infinitely thin and infinitely numerous. It's still an accumulation of tiny pieces — just carried all the way to the limit, so the leftover gaps vanish completely.

$a, b$The left and right edges of the region being measured
$f(x_i)$The height of the curve at the $i$-th rectangle's sample point
$\Delta x$The width of each thin rectangle
$\lim_{n\to\infty}$Letting the rectangle count grow without bound, so the width shrinks to zero

Think Further

  1. In the worked example, the right-endpoint sum (10) was larger than the exact area (8). Would using left endpoints instead give a sum that's too big or too small for this increasing function? Why?
  2. An autonomous vehicle only has speed readings sampled once per second, not a continuous speedometer curve. How does a Riemann sum let it still estimate total distance traveled?
  3. If a curve dips below the x-axis for part of the interval, what do you predict happens to the "area" contributed by that portion of the Riemann sum?
Show suggested answers
  1. Too small. Since $f(x)=x$ is increasing, the left edge of each rectangle sits below the curve, so left-endpoint heights underestimate, giving a sum smaller than the true area of 8.
  2. Each sampled speed reading becomes the height of one rectangle, and the one-second gap becomes its width. Multiplying and summing those speed×time rectangles approximates total distance, exactly like a discrete Riemann sum.
  3. Rectangles below the x-axis have negative height, so they contribute negative area to the sum. The Riemann sum treats area below the axis as subtracting from the total rather than adding to it.