Posts: 82

Threads: 15

Joined: Jan 2008

" 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

Posts: 253

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Joined: Jun 2012

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

Posts: 3,283

Threads: 104

Joined: Jul 2005

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.

Posts: 50

Threads: 1

Joined: Sep 2011

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