Area under a curve, Archimedes' way
Tonight's idea
Straight edges are easy; curves are the problem. Take the region under \(y = x^2\) between \(x = 0\) and \(x = 1\). Tonight you find its area in three steps.
Slice \([0, 1]\) into \(n\) equal strips of width \(\tfrac{1}{n}\). On each strip stand a rectangle whose height is the curve's value at the strip's right-hand edge. The \(k\)-th strip ends at \(x = \tfrac{k}{n}\), so its rectangle has area \(\tfrac{1}{n}\cdot\tfrac{k^2}{n^2}\) (the figure shows \(n = 4\)). Sum them, using the sum-of-squares formula (provable by induction):
\(R_n = \dfrac{1^2 + 2^2 + \cdots + n^2}{n^3} = \dfrac{n(n+1)(2n+1)}{6n^3} = \dfrac{1}{3} + \dfrac{1}{2n} + \dfrac{1}{6n^2}.\)
Now squeeze. Right-hand rectangles poke above the curve, so \(R_n\) is too big. Left-hand edges give \(L_n = \tfrac{1}{3} - \tfrac{1}{2n} + \tfrac{1}{6n^2}\), too small. The true area \(A\) is trapped: \(L_n \lt A \lt R_n\) for every \(n\). As \(n\) grows both bounds close in on \(\tfrac{1}{3}\), and the only number between them forever is \(\tfrac{1}{3}\). So \(A = \tfrac{1}{3}\). Exactly. This is the integral; Archimedes did it around 250 BC. Play with the slices at mathdetours.com/demos/riemann.html.
Watch one: four slices
With \(n = 4\) the right-hand heights are \(\tfrac{1}{16}, \tfrac{4}{16}, \tfrac{9}{16}, \tfrac{16}{16}\), each on a base of \(\tfrac14\), so \(R_4 = \tfrac14 \cdot \tfrac{30}{16} = \tfrac{30}{64} = 0.46875\). The left-hand heights are \(0, \tfrac{1}{16}, \tfrac{4}{16}, \tfrac{9}{16}\), so \(L_4 = \tfrac{14}{64} = 0.21875\). The gap is \(\tfrac{16}{64} = \tfrac14\), exactly \(\tfrac1n\): the overhanging pieces slide together into one column of width \(\tfrac1n\) and height 1.
Your turn
- Compute \(L_5\) and \(R_5\) for \(y = x^2\) on \([0, 1]\) by hand, as fractions. Check that \(\tfrac13\) sits between them and that they differ by \(\tfrac15\).
- Run the same method on \(y = x\) over \([0, 1]\), using \(1 + 2 + \cdots + n = \tfrac{n(n+1)}{2}\). Simplify \(R_n\), say what it tends to, and check it against the area of a triangle.
- A gift: \(1^3 + 2^3 + \cdots + n^3 = \left(\tfrac{n(n+1)}{2}\right)^2\). Use it to find the area under \(y = x^3\) on \([0, 1]\). Look at the answers for \(x\), \(x^2\) and \(x^3\) together and guess the area under \(x^{10}\).
- Now \([0, 2]\) under \(y = x^2\). The strips have width \(\tfrac{2}{n}\) and the \(k\)-th right-hand height is \(\left(\tfrac{2k}{n}\right)^2\). Show the area is \(\tfrac83\), and then find the area from 1 to 2.
Stretch: the area under 1/x
Call the area under \(y = \tfrac1x\) from \(1\) to \(2\) by the name \(A\). Stretch the picture horizontally by a factor of 2 and squash it vertically by a factor of 2. What happens to areas? To the curve? Where does the region over \([1, 2]\) land? Find the area from \(1\) to \(4\) in terms of \(A\), with no arithmetic.
Answers
- \(R_5 = \tfrac{1}{125}(1 + 4 + 9 + 16 + 25) = \tfrac{55}{125} = \tfrac{11}{25}\). \(L_5 = \tfrac{1}{125}(0 + 1 + 4 + 9 + 16) = \tfrac{30}{125} = \tfrac{6}{25}\). \(\tfrac{6}{25} \lt \tfrac13 \lt \tfrac{11}{25}\), and \(\tfrac{11}{25} - \tfrac{6}{25} = \tfrac15\).
- \(R_n = \tfrac{1}{n^2}(1 + 2 + \cdots + n) = \tfrac{n+1}{2n} = \tfrac12 + \tfrac{1}{2n} \to \tfrac12\). The region is a right triangle with legs 1 and 1, area \(\tfrac12\).
- \(R_n = \tfrac{1}{n^4}\left(\tfrac{n(n+1)}{2}\right)^2 = \tfrac{(n+1)^2}{4n^2} \to \tfrac14\). The pattern \(\tfrac12, \tfrac13, \tfrac14\) says the area under \(x^m\) on \([0, 1]\) is \(\tfrac{1}{m+1}\); for \(x^{10}\), \(\tfrac{1}{11}\). (It is true.)
- \(R_n = \tfrac{2}{n}\sum \tfrac{4k^2}{n^2} = \tfrac{8}{n^3}\sum k^2 = 8\left(\tfrac13 + \tfrac{1}{2n} + \tfrac{1}{6n^2}\right) \to \tfrac83\). From 1 to 2: \(\tfrac83 - \tfrac13 = \tfrac73\).
Stretch. The stretch-and-squash keeps every area the same (widths double, heights halve). The point \((x, \tfrac1x)\) goes to \((2x, \tfrac{1}{2x})\), still on the curve, so the curve maps onto itself and the region over \([1, 2]\) lands exactly on the region over \([2, 4]\). So the area from 2 to 4 is also \(A\), and the area from 1 to 4 is \(2A\). This area behaves like a logarithm, which is exactly what \(\ln\) is: \(A = \ln 2 \approx 0.693\).
The next surprise: Area and slope undo each other. Move the edge of an area and watch its slope become the curve's height.