Integral of floor(x)*(x)*ceiling(x) dx from 0 to 4

[@weo9473]

Nah bro you're obsessed with floor and ceiling functions

[@NetheriteMinerMicrowavegang]

I don't see the problem with that if it means he keeps putting out banger videos🔥🔥🔥

[@Edsonrsmtm]

Kk😂😂

[@cadenpink316]

Me too

[@treybell40501]

It would lowkey b funny now if he drops a video about one each week now lol

[@Nzargnalphabet]

Why not though?

[@Christian_Martel]

7:50 For the purpose of generality, the index of the sum starts at k = 0 and goes to k = n - 1. The first term will be 0.

[@JWheadset]

We can actually further extend this to negative integers as well. In this case, the index of the sum starts at k = a and ends at k = b - 1, given that we integrate from x = a to x = b.

[@AbouTaim-Lille]

This is not a continuous function but it is integrable by Lebesgue. We can write x = t+n. Where 0

[@davidhowe6905]

You can simplify this further using formulae for sums of integers, squares and cubes, which simplifies nicely (using n as the upper integration limit i.e. n=4 in the example) giving (n^4 - n^2)/4 (I think this is right)

[@kujmous]

Loved the approach of getting an answer and then getting a general form. This was right on the cusp of my abilities, and I appreciate it very much. Well done.

[@Mini_Wolf.]

Hi, love your vids, at 12:00 when it was six times the first squares how would you go from there? would you add another summ?

[@PrimeNewtons]

Yes. It becomes a sum of sums. I'll do a video soon.

[@Mini_Wolf.]

@@PrimeNewtons oh okay, ty, I'll be the first to watch it

[@pedrogarcia8706]

fantastic video! at 15:45 instead of leaving that middle fraction as 3/12 though you could have reduced it to 1/4

[@m.h.6470]

yes, and then - with the correction at the end - all terms have the same denominator and can be combined to (k⁴ + 4k³ + 5k² + 2k)/4 or k/4 * (k³ + 4k² + 5k + 2)

[@rainerzufall42]

@@m.h.6470 Brilliant remark! Never stop looking for patterns! It's even more easy... ( (k(k+1))² + k(k+1)(2k+1) + k(k+1) ) / 4 = ( k * (k+1)²*k + (2k+1) * (k+1)k + 1 * (k+1)k ) / 4 = ( k * (k+1)²*k + (2k+2) * (k+1)k ) / 4 = ( k * (k+1)²*k + 2 * (k+1) * (k+1)k ) / 4 = ( (k+2) * (k+1)²k ) / 4 = k(k+1)(k+1)(k+2) / 4. With n = k + 1, the integral is I(n) = n²(n²-1)/4. For k = 3 => n = 4: int_0^n floor(x) * x * ceil(x) dx = n²(n²-1)/4 = 16 * 15 / 4 = 240 / 4 = 60. For n = 100: I(100) = 10.000 * 9.999 / 4 = 99.990.000 / 4 = 24.997.500 - Pretty easy, ey? No need for a calculator!

[@rainerzufall42]

Brilliant remark! Never stop looking for patterns! It's even more easy... ( (k(k+1))² + k(k+1)(2k+1) + k(k+1) ) / 4 = ( k * (k+1)²*k + (2k+1) * (k+1)k + 1 * (k+1)k ) / 4 = ( k * (k+1)²*k + (2k+2) * (k+1)k ) / 4 = ( k * (k+1)²*k + 2 * (k+1) * (k+1)k ) / 4 = ( (k+2) * (k+1)²k ) / 4 = k(k+1)(k+1)(k+2) / 4. With n = k + 1, the integral is I(n) = n²(n²-1)/4. For k = 3 => n = 4: int_0^n floor(x) * x * ceil(x) dx = n²(n²-1)/4 = 16 * 15 / 4 = 240 / 4 = 60. For n = 100: I(100) = 10.000 * 9.999 / 4 = 99.990.000 / 4 = 24.997.500 - Pretty easy, ey? No need for a calculator!

[@rainerzufall42]

My answer is invisible unless you change the sort order!

[@doctorb9264]

Excellent problem and approach.

[@LearnLife17]

Just came across your channel watched 1 video and the way you explain is so great. Can you kindly recommend some great math books

[@NLGeebee]

7:22 classic maths teacher move 😂

[@dimsergeyev]

I love that guy! Superb! Thank You a lot!

[@rainerzufall42]

Short calculation (see below!): I(n) = int_0^n floor(x) * x * ceil(x) dx = n² (n²-1) / 4. Makes things easier! n = 4: I(4) = 16 * 15 / 4 = 240 / 4 = 60.

[@MASHabibi-d2d]

Thanks, can you tell me what the formula of sum of sentences is for higher powers, for example k^5

[@FernandoRamirez-sg9ll]

Good job! 👍

[@Dalroc]

Interval 0->1: floor is 0, so whole integral is 0. Interval 1->2: floor is 1, ceiling is 2, so function becomes 2x and integral is x². Interval 2->3: floor is 2, ceiling is 3, so function is 6x and integral 3x². Interval 3->4: floor is 3, ceiling is 4, function is 12x, integral 6x². First integral: 0 Second integral: 2²-1² = 3 Third integral: 3(3²-2²) = 15 Fourth integral: 6(4²-3²) = 42 0+3+15+42 = 60 Answer is 60.

[@spoopy1322]

Amazing and super interesting!❤ I'm wondering how you would go on to solve the indefinite integral of this function

[@PrimeNewtons]

I don't know.

[@carstenmeyer7786]

@@PrimeNewtons You get a piece-wise quadratic: *n < x

[@vaibhavsrivastva1253]

It much simpler to just write floor(x) = {0 if 0

[@vitotozzi1972]

Simply awesome!!

[@oshintimothy6042]

Archsin((8x)/(16+x²))dx please help me solve

[@tungyeeso3637]

When there's 100 terms, we have to do 100 different additions. Is there an easy way out? The tedious addition is going to kill anybody. 😅

[@vanouper]

Thumbnail: Integral of floor(x) xceiling(x) dx from 0 to 4 Title: Integral of floor(x) xceiling(x) dx from 0 to 3 Just wanted to let you know.

[@nooooo3932]

the title says that the integral is from 0 to 3 but the thumbnail says that the integral is from 0 to 4...

[@m.h.6470]

Solution: ₀∫⁴ ⌊x⌋x⌈x⌉ dx Just by looking at the boundaries, we can move the boundaries from "0 to 4" to "1 to 4", as absolutely nothing happens between 0 and 1, as ⌊x⌋ = 0. For 1 ≤ x < 2, we have the equation 1 * x * 2 = 2x. For 2 ≤ x < 3, we have the equation 2 * x * 3 = 6x. For 3 ≤ x < 4 we have the equation 3 * x * 4 = 12x. So we have: ₀∫⁴ ⌊x⌋x⌈x⌉ dx = ₁∫² 2x dx + ₂∫³ 6x dx + ₃∫⁴ 12x dx = [x²]₁² + [3x²]₂³ + [6x²]₃⁴ = 2² - 1² + 3*3² - 3*2² + 6*4² - 6*3² = 4 - 1 + 27 - 12 + 96 - 54 = 3 + 15 + 42 = 60

[@mscha]

You really shouldn't have re-used 𝑘 like that, that's confusing. Better to use 𝑛. By the way, if you clean this up further, you end up with a pretty nice formula: 𝑛²(n²-1)/4 (where 𝑛 ∈ ℕ is the top of the integral, e.g. 4 in this case).