Wikipedia:Reference desk/Archives/Mathematics/2009 December 8

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December 8[edit]

Standard form[edit]

Resolved: Accdude92 (talk to me!) (sign) 15:16, 8 December 2009 (UTC)

I found out how.

How do you put y=-4/3x-8 into standard form? You need to get everything to be a whole number, but i don't know what to multiply everything to do that.Accdude92 (talk to me!) (sign) 14:51, 8 December 2009 (UTC)

~~ Dr Dec (Talk) ~~ 15:11, 8 December 2009 (UTC)
More likely . Then multiply everything by 3 to get something like what you described. Staecker (talk) 15:14, 8 December 2009 (UTC)

Geometric argument for area under x2[edit]

Does anybody know an elementary geometric argument (slicing and reattaching or the like) that shows that the area under x2 from 0 to 1 is 1/3? I tried for a good 20 minutes and couldn't come up with anything. Equivalently you could try to show the area between x2 and is 1/3, which doesn't seem any easier. Staecker (talk) 15:17, 8 December 2009 (UTC)

Archimedes investigated this problem in The Method of Mechanical Theorems and found that the area is 1/3. He used an argument that imagined the parabola on a lever, balancing a triangle on the other side of the fulcrum. I don't know if this is as "elementary" as you want—it does use the idea of infinitesimals—but it's different from the typical calculus approach at least. —Bkell (talk) 15:26, 8 December 2009 (UTC)
Interesting- certainly not something I've seen before. Not really as elementary as I wanted, but I suspect that I'm asking for too much. Staecker (talk) 15:54, 8 December 2009 (UTC)
The standard way of deriving the definite integral of from the definition of Riemann integral uses the identity The (upper) Riemann sum for with respect to the uniform partition is
Since the above Riemann sum is as so the integral is 1/3. --pma (talk) 20:40, 8 December 2009 (UTC)
PS: As far as I recall Archimedes computes the area of a segment of parabola by the exhaustion method, filling it with triangles; he has a simple lemma about the area of an inscribed triangle, and then the final result is equivalent to summing a geometric series with ratio 1/4. --pma (talk) 20:57, 8 December 2009 (UTC)
Also, you can interpret the integral of x2 on [0,1] as the volume of a pyramid with unit square base and height 1. --pma (talk) 21:30, 8 December 2009 (UTC)
Thanks pma- I'm familiar with Riemann sums. (I taught it today to my freshman calc class.) The pyramid volume sounds close to what I'm looking for, but I can't see how that works. Can you explain more? Staecker (talk) 22:06, 8 December 2009 (UTC)
These two links describe the process in detail - Essay on Quadrature of Parabola, Clear Photos of Translation of Archimedes Original Proof. Hope this helps. --PST 01:45, 9 December 2009 (UTC)
OK, that's pma's "PS" method. Those links are good, but this is still much more complicated than what I was looking for (still using an infinitesimal-type argument in the summing). I guess maybe there just isn't a simple argument. But I was intrigued by pma's comment about the volume of a pyramid. Are you (PST) saying that your links explain that too? (If so I need to read them more carefully because I didn't see anything about pyramids.) Staecker (talk) 02:21, 9 December 2009 (UTC)
Does the bottom of this page have what you are looking for? --PST 02:38, 9 December 2009 (UTC)
Explanation of pyramid volume: Consider a pyramid with unit square base and height 1. It's made of infinitesimal cubic bricks. Count them layer by layer, numbering each layer by its distance from the tip. We find that the area in layer x is x2, and the volume of the entire pyramid is therefore . But that is exactly the same as the area under the parabola you're asking for. You probably need to appeal to infinitesimals to make this identification, but it should be possible to derive the volume of the pyramid without infinitesimals. –Henning Makholm (talk) 03:02, 9 December 2009 (UTC)


Pyramid of Khafra
Agreed. To identify the volume of the pyramid (on the left) with the integral of x2 one may use the following exhaustion argument. First interpret the upper and lower Riemann sums as the volume of the outer and inner terraced pyramids (on the right); this equality volume=Riemann sum is a direct verification. Then, we compute the difference between the volume of the inner and outer terraced pyramids with layers of width 1/n, and note that it is exactly 1/n (it is made by n co-centric square frames that put together form a rectangular parallelepiped of size 1 × 1 × 1/n), which allows to extend the equality volume = integral to the Egyptian fashion pyramid. --pma (talk) 11:27, 9 December 2009 (UTC)
Great- thanks for those descriptions. I see it now. (Though I still don't see it in PST's link.) So it still relies on infinitesimals, but in a much simpler way than the two Archimedes methods referenced above. I guess that's as simple as it's going to get. Thanks- Staecker (talk) 12:48, 9 December 2009 (UTC)


Talking about buildings, there's also a nice way to compute the sum of the first n squares. Consider 6 rectangular parallelepipeds of size 1 × k × k, (you can think each of them as made by k2 unit-cube bricks), and arrange them to form the walls of a room (with no floor and no ceiling) of outer size (2k+1) × (k+1) × k and inner size (2k-1) × (k-1) × k. Note that the inner size of the k-th object coincides with the outer size of the (k-1)-th object. Therefore, you can put a collection of them, from k=1 to n, one inside the other like matryoshka dolls and form a full parallelepiped of size (2n+1) × (n+1) × n, which explains the formula.--pma (talk) (talk) 11:27, 9 December 2009 (UTC)


Here's another more algebraic argument, that may be of interest for your class. Call α the integral of x2 over [0,1]. The integral over [0,2] is then 8α: you may justify it to your class by considering the linear transformation (x,y)↦(2x,4y), that expands all areas by a factor 8 (that's true for rectangles, hence also true for the area of the parabola), or alternatively, by a linear change of variable formula for the Riemann integral, to be proven separately. But the integral of x2 over [0,2] is also the sum of the integral over [0,1] plus the integral over [1,2]; the latter may be written as the integral of (x+1)2 over [0,1], and expanded by linearity of the integral into a sum of three integrals. Assuming they already know what's the integral of x and of 1, we end up with the linear equation 8α=2α+2 (I didn't try to solve it but I'm quite confident that it will give α=1/3 ).--pma (talk) (talk) 13:01, 9 December 2009 (UTC)
Thanks- that's a nice argument. Staecker (talk) 14:41, 9 December 2009 (UTC)
Here's an elementary argument, that turns the integral into a geometric problem. I'll write it out in a lot of detail, but many steps are obvious and can be skipped.
We can interpret the integral of as half of the center of mass of the line from 0 to 1 where the point at position x has mass x dx:
Equivalently, M can be interpreted as the x-coordinate of the center of mass (i.e., the centroid) of the triangle T with vertices at (0, 0), (1, 0), and (1, 1). So it remains to be shown that the x-coordinate of the centroid of T is 2/3.
Unfortunately the direct way to compute the centroid of T involves an integral, which just gets us back to where we started. However, by appealing to physical intuition, one can argue that the centroid lies on the medians of a triangle (use the fact that the area of a triangle is half of base times height and maybe some hand-waving). Thus one needs to show (the well-known fact) that the intersection of the medians divides the medians into segments whose lengths have ratio 2:1. One way to do that is, show that it is true for equilateral triangles (divide an equilateral triangle into its medians and use similar triangles), and then argue that the ratio 2:1 is preserved by linear transformations (scaling, skewing, rotating), and that any triangle can be gotten from an equilateral triangle via linear transformations (just skew the sides until they're all the right length).
Hopefully the above is clear; most of it should generalize to higher dimensions, too. (I put a diagram of the linear transformation thing here.)
Incidentally, question B2 on the Putnam exam last week boiled down to computing upper Riemann sums for .... Eric. 131.215.159.171 (talk) 01:39, 10 December 2009 (UTC)
Thanks- that's basically the same as the method at The Method of Mechanical Theorems by Archimedes. Staecker (talk) 13:39, 10 December 2009 (UTC)
Oh -- that is very similar. Cool. Eric. 131.215.159.171 (talk) 20:10, 11 December 2009 (UTC)