Mathematical limerick

I may have posted this before, but even if I did, it deserves a repost.




The integral of zee-squared dee zee,
From 1 to the cube root of 3,
Times the cosine,
Of 3 pi by 9,
Is the log of the cube root of e.


What's even better is, it's accurate. I've confirmed this myself, but I have it from reliable sources...

Hah, that's awesome. And it does indeed appear correct.

That's a pretty picture you have there....... lol

*Head explodes*

I can see how the limerick relates to the picture but I suck at maths

Yes, it is definitely correct.

The integral of z^2 is (z^3)/3, and to evaluate an integral you take the integral evaluated at the upper value and subtract from it the integral evaluated at the lower value.


So you get:

((3^1/3)^3/3 - (1)^3/3) * (cos(pi/3))

(3/3 - 1/3) * (1/2)

(2/3) * (1/2)

= 1/3

The ln of e raised to any power is just the value of the power, since that's literally the definition of the ln function. So, e^(1/3) is just 1/3.

1/3 = 1/3

Impressive.