/sci/ - Science & Math

its actually just 0 + c

>>5056205
Not possible.
The derivative of every differentiable function is Darboux continuous but floor(x) isn't.

Intuitively. Consider that floor(0) = 0. floor(0.5) = 0, floor(1) = 1, floor(1.5) = 1, etc. The integral can be viewed as the signed area under a curve. If we assume x > 0, floor(x) = 1 + 2 + ... + x - 1. Therefore, Integral floor(x)dx from a to b is Sum[i=a,b] i.

>>5056216
so then wouldn't the integral be:
floor(x)*x?

Did I solve the age old questions guys? Where us my prize?

There has to be a pattern here, there has to be a way to solve for any floor(f(x))

>>5056216
That should be Sum[i=a,b-1].

>>5056224
Possibly, but it wouldn't be in terms of standard mathematical functions, I would imagine.

>>5056216
>>5056228
this, except it is not [i=a, b-1] but [i=a, floor(b) - 1]

>>5056216
>If we assume x > 0, floor(x) = 1 + 2 + ... + x - 1
I worded this in the worst way. What I should have said was, if we assume x > 0 and we're working on the interval [0,x], the area under the curve floor(x) = 1 + 2 + ... + x - 1.

>>5056239
you even can put this in a closed formula:
[floor(b)*(floor(b)-1)]/2

>mfw people refuse to accept laws of physics.

floor a (ceiling a - a) +
floor b (b - floor b) +
sum (ceiling a, floor b) n

That's the definite integral.

>>5056255
>thinking that a picture is related unless specified on /sci/
mfw

>>5056338
Sorry, I was being distracted.
The definite integral of floor(x) is:
{sum(i = floor(a), floor(b)) i} -
{floor(b) (ceiling (b) - b)} -
{floor(a) (a - floor (a))}