integral of floor of x from 0 to 4 (greatest integer function)

Here we will integrate floor(x) from 0 to 4. The floor function is also called the greatest integer function.

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[SlimThrull]

so what's the answer from 0 to infinity? Is it infinity or -1/12?

[sdegueldre]

A more general answer would've been nice, like an indefinite integral. would be something like x²-(x-⌊x⌋) right?

[maccollo]

floor(x⌋ actually has a "primitive function", in the sense that it is continuous and it's right side derivative matches the value of floor x.
The function is: F(x) = x⌊x⌋ - ⌊x+1⌋⌊x⌋/2 + C.

You can find expressions for other integrals of functions like this, like ⌊sin(x)⌋ works, and also ⌊log(x)⌋, and ⌊sqrt(x)⌋.
The trick to solving them is that when you integrate you don't just get a constant C, but a function R -> Z which plugs in the discontinuities you get. There's a bit of a trick to figuring it out where you look at the sum of the discontinuities.
There are also some that are impossible (correct me if I am wrong) like a primitive function of ⌊x²⌋, because the sum that pops up is the sum of square roots.

[leastmachine]

I know I'm two years late, but why exclude 4, the upper bound of the integral? Also, why integrate at all when the function is discontinuous over the interval?

[swizzbeats]

Ah, great video. Never realized how easy of a concept the floor function is.

[AndurCarr]

After thinking about this and the related problem ( ∫ {x} dx ), I think the floor should be defined with the fractional part instead of the other way around:
⌊x⌋ = x - {x}

For the anti-derivative of the fractional part:
∫ {x} dx = (x - {x} + {x}^2)/2 + C

Therefore:
∫ ⌊x⌋ dx = ∫ x dx - ∫ {x} dx = (x^2)/2 - (x - {x} + {x}^2)/2 + C

[martind]

Integrating the floor function is super easy, barely an inconvenience.

[mihaly]

so how would you do these if it’s like any function inside the floor. is there a general method of attacking these questions

[SasquatchlifeX]

LOL! I had a feeling you would use simple geometry!

[waynemv]

Easy. You should have made it a very slight bit harder by doing the integral from 3 to 3 of floor(x) dx. Or more, generally, from a to b of floor(x) dx.

[orphixigl]

You can prove by induction that if n is a natural number
Integral 0, n ([x]) = n(n-1)/2
If a and b are real numbers and 0≤a≤b then
Integral a, b ([x])=
= [b][b-1] - [a][a-1] + [b](b-[b]) - [a](a-[a])

[Ynook]

You can still do that by integration. Integral of floor(x) is (x^2-x)/2. In your case, (4^2-4)/2 - (0^2-0)/2 = (16-4)/2=6.

[Mathematcs-]

Nice ) saw this function in nuberfile channel lol

[justabunga]

On your graphing calculator, the greatest integer or floor function is denoted as int. For ceiling or least integer function, you don’t have it on your graphing calculator. Instead you would have to type it in as -int(-number). e.g. ceil(x)=-int(-x). You would type this integral function as fnInt(int(x), x, 0, 4), which is 6.

[Dunklesteus]

The explanation is drawn out like it's for little kids but it's an integral... who is the target audience for this video??

[i_am_anxious]

I saw that intro and I had two reactions;

“No YAY?”

And

“Pmmff. This is gonna be good.”

It was, noice vid dude!

[plasmaballin]

Easy if you know how triangular numbers work. The indefinite integral is just Δ⌊n-1⌋+⌊n⌋⌈n⌋+C, where Δn is n(n+1)/2 and ⌈n⌋ is the fractional part of n.

[calyodelphi]

Interesting thing about integrals of the floor and ceiling functions is that if the bounds of integration are integers, they are just summations in disguise:

integral from a to b of floor(x) dx = summation from x = a to b-1 of x
integral from a to b of ceiling(x) dx = summation from x = a+1 to b of x

The integral of the nearest integer function is a bit more difficult though...

integral from a to b of round(x) dx = summation from x = a to b of x, but that gives you a bit of excess on the ends. You then have to *subtract* half of a and b from the summation in order to properly chop off those extra ends. So it comes out to be [sum from x=a to b of x] - (a+b)/2.

All three of these are conditional that a, b are in Z.

If a and b are in R, then things get a bit more dicey, because in order to do the summation, you have to first round a and b up/down, respectively, do the summation there, then a little bit of extra math to account for the extra bits on the ends.

I'm mostly just writing this off the top of my head right now and haven't taken the time to really generalize it yet, but yeah. I believe Dr. Peyam covered this in a previous video as well, and I was pleased as punch with myself when I paused at the beginning and thought it through for half a moment and recognized it for what it was. ^_^

[juliengrijalva]

this yields a fun way to find the formula for the sum of the first n natural numbers

[meerable]

Is the "floor" function
continuous? Is it possible to integrate it at all? Or do we need some reservations?