The answer to the second semi-mathematical puzzle

[Mike Reed]

" int_(1)^(3^(1/3)) z^2 dz cos(3pi/9) = ln e^(1/3)

(Note: Readers outside the US, this works better if you allow yourself the agony of pronouncing the letter “z” as “zee”.)

I will post this answer one week from today"

OK, the week has passed; here's the answer:

The integral of z squared dz
from 1 to the cube root of 3
times the cosine
of 3 pi over 9
equals log of the cube root of e

How did you do in solving it?

mike

[Les Koller]

Not sure how to read it in Lymerick meter...is this close?

The integral of z squared dz
from one to cube root of 3
times the cos of 3pi
divided by nine
= the log of the cube root of three

[Marcus von Cube, Germany]

The integral of z squared dz

from one to cube root of 3

times the cos of 3pi

divided by nine

= the log of the cube root of three

The formum software has wrapped the text. What you see here is enclosed in [pre]...[/pre] tags. As an alternative use [nl] at each line end.

[Derek Walker (UK)]

Shouldn't 'cosine' and 'nine' be at the ends of lines, to rhyme:


The integral of z squared dz
from 1 to the cube root of 3
times the cosine
of 3 pi over 9
equals log of the cube root of e