Riemann sums

What you are looking at

A Riemann sum approximates $\int_a^b f(x)\,dx$ with rectangles. Each strip has width $\Delta x = \dfrac{b-a}{n}$ and a height $f(x_i)$ taken at its left edge, its right edge or its midpoint; with 40 strips or fewer, the sample points are marked on the curve. The working panel shows $\Delta x$, the sum $\sum_{i=1}^{n} f(x_i)\,\Delta x$ and its value, and the exact integral from an antiderivative, $\big[F(x)\big]_a^b$, so the two numbers sit side by side.

The definite integral is the limit of these sums as $n \to \infty$. Press Increase $n$ and $n$ steps up towards 100 while the rectangles hug the curve more closely. The Difference section plots the size of the gap between the sum and the integral for every $n$ from 1 to 100, with a marker at the current $n$. Rectangles below the $x$-axis count as negative, so the sums approach the signed integral, not the area.

Try this

1. 1Leave $f(x) = x^2$ on $[0, 2]$ and press Increase $n$: the left sum climbs from about $2.04$ at $n = 6$ towards $\dfrac{8}{3} \approx 2.667$.
2. 2Switch between Left, Right and Midpoint at the same $n$: for this increasing curve the left sum is too small, the right sum too big, and the midpoint sum far closer than either.
3. 3Choose $\sin x$, where $[0, \pi]$ gives an exact value of 2, then drag $b$ to $2\pi$: the exact value drops to 0, because the part below the axis counts as negative.
4. 4Press Hide formulas, decide whether right rectangles overestimate or underestimate the integral, then show the formulas again and check the Difference section.

Mistakes students make

• Believing left sums always underestimate. That holds only when $f$ is increasing on $[a, b]$; where $f$ decreases the left sum is too big, and on a curve that rises and falls it can go either way.
• Getting the sample points wrong. With $\Delta x = \dfrac{b-a}{n}$, left sums use $x_i = a + (i-1)\Delta x$ and right sums $x_i = a + i\,\Delta x$ for $i = 1, \dots, n$: $n$ heights, not $n + 1$.
• Treating a Riemann sum as an area. Rectangles below the $x$-axis contribute negative amounts, so the sums approach the signed integral: for $\sin x$ on $[0, 2\pi]$ that is 0, although the area is 4.

Where it is taught

• IB Maths AA SL/HL 5.5 · 5.11
• AP Calculus AB/BC Unit 6
• Edexcel A-Level Pure Year 2 · Integration as the limit of a sum
• Greek Lykeio year 3 (Γ΄ Λυκείου) · §3.4