[Arc length = area]

If the function was a straight line, you could use the Pythagorean theorem to find it.

sqrt( (y2-y1)^2 + (x2-x1)^2 )

If you assume the arc length was a straight line, and slowly bring the points on the function closer together, eventually you get the change in y approaching the slope of the line times the difference in x's.

The arc length for that infinitesimally small section is then sqrt( ( f'(x)dx )^2 + dx^2 ) = sqrt( dx^2 * (f'(x)^2 + 1 ) = sqrt(   f'(x)^2 + 1 ) dx.  Where f'(x) is the derivative of f(x).

Integrate that between two points to get the arc length between those two points.

Example: f(x)=2x

f'(x) = 2

integrate( sqrt( 2^2 + 1 ) dx ) = integrate( sqrt(5) dx) = x*sqrt(5) + c

If you evaluate the integral between 0 and 1 you end up with arc length =  sqrt(5).  Which is the same as if you just used the Pythagorean theorem.  Yeah, I'm too lazy right now to find a function that ends up with a nice arc length function that isn't just a straight line

[If then puzzle]

if 12=6 , 6=3 10=what

answer is 10=5 because 

6 x 2 = 12

3 x 2 = 6

its just half of the number x 2 example

half of 12 = 6 x 2 = 12