Integral Battle! With Floor Functions
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- Опубликовано: 14 янв 2019
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Integral battle between integral of (floor(x))^2 vs. integral of floor(x^2)
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For the second integral, instead of using those really long and skinny rectangles with widths along the x-axis, it would be easier to take the 60 as equal to the full x domain (4) times the full y range (15), then subtract out rectangles with widths along the y axis. There are 15 of them, they each have width 1, and they each have length sqrt(n), for n from 1 to 15.
Congratulations for 200k subscribers! Awesome content as always, love your channel
Blackpenredpen you have 200K subs! Congrats!
you can hear people cheering at 1:50
Thanks!!!!!
But it is deserve 50 million subs
Now he has 750K
1. Consider the integral from a to b of f[floor(x)]. You can use the connection rule of integration to separate this integral into three integrals, from the intervals a to floor(a), floor(a) to floor(b), and floor(b) to b. Do make sure that a < b for simplicity, which can always be ensured, because if you have an integral from b to a when b > a, then you can switch the order of integration by multiplying by -1. Then evaluate each separately, which merely involves taking the area of rectangles. In particular, the integral from floor(a) to floor(b) is equal to the sum of f(n) from n = floor(a) to n = floor(b) - 1. The integral from a to floor(a) is simply f[floor(a)]·[floor(a) - a]. Depending on your definition of {x}, the latter could simply be -f[floor(a)]·{a}. Meanwhile, the integral from floor b to b is simply f[floor(b)]·[b - floor(b)] = f[floor(b)]·{b}. Combining all three gives that the integral from a to b of f[floor(x)] is (f[floor(b)]·{b} - f[floor(a)]·{a}) + S[floor(a), floor(b) - 1; f(n)], where S[floor(a), floor(b) - 1; f(n)] is the sum from n = floor(a) to floor(b) - 1 of f(n). This works for any f and any interval of the real numbers.
Yo! Congrats on 200k subs 1/(1-x), you deserve it man! Actually, no. You deserve way more than that.
Pi is the best
Thank you!!!!
blackpenredpen you’re awesome man, love your vids! Keep up the great work!
2. Consider the integral of floor[f(x)] from a to b, with respect to x. If f is piecewise continuous in the interval, then there exists some set {c_1, c_2, ..., c_k} contained in the interval (a, b), possibly the empty set, such that for all ε > 0, f(c_i - ε) < n_i & f(c_i + ε) > n_i, where n_i indexes over a set of integers in the range of f.
If this set is non empty, then you can consider the integrals from a to c_1 of floor[f(x)] + integral from c_1 to c_2 floor[f(x)] + ••• + integral c_(k - 1) to c_k floor[f(x)] + integral c_k to b floor[f(x)]. Now each integró is trivially given, since floor[f(x)] is a constant in each interval, though a different one in each interval. An integral from c_i to c_(i + 1) of floor[f(x)] is equal to (n_i)[c_(i + 1) - c_i], where n_i is the corresponding integer (from the set of indexed integers I described above) to c_i as outlined in the epsilon-inequalities I stated. Combining all such integrals gives (n_1)(c_2 - c_1) + (n_2)(c_3 - c_2) + ••• + [n_(k - 1)][c_k - c_(k - 1)] = -(n_1)(c_1) - (n_2 - n_1)(c_2) - ••• [n_(k - 1) - n_(k - 2)][c_(k - 1)] + [n_(k - 1)]c_k. Meanwhile, from a to c_1 you have floor[f(a)][c_1 - a], and from c_k to b you have n_k(b - c_k).
If the set is empty, then floor[f(a)] = floor[f(b)] = M. Then the integral is simply M(b - a).
That's a 600k+ subscribers ! Congratulations. I would have drawn the parabola y=x^2 for the second integral.
200K subs! Looks like you will make it to 400K before the end of the year
Also, the link on the description only gives the sum of the first n squares, not the first n square roots, or so it appears. You could define a recursive formula, though, I think
Thanks Angel!!
I think I can hit 300k without problem but 400k is a bit challenging.
Awesomely done!!!
Will you ever be doing a video on Lagrange's multipliers?
He made a video on it last year.
ruclips.net/video/ch4PPp_bQtI/видео.html
I think a good advice if you want to expand beyond 400k subs is by going on multiple social media websites . Don't stop at youtube. Go to Facebook, twitter , tumblr etc. . And also a good idea is to translate your video into more languages so your content can be seen by people from multiple countries . However, my advice is
theoritical,not sure how it's going to work practically
Hi Kimo, thanks for the comment.
I am already on twitter but not fb. The translation idea will work however it will take a tremendous amount of time (or money) of me to do so. I might be able to do some for some of my popular videos but certainly not all. Thanks again for your input.
Your welcome and my actual name is karim
kimo the fun genius thanks karim
I check on geogebra for this integral and it's match with ur answer.
Just YAY!
First integral is 14 and the second is 19,52
Congratulations! on 200k subscribers!!!!! yay...yay...yay
Hello, I have a question hope you can make a video about.. the domain an range of (y = (3 - sin(x))^0.5)
Can you make worksheets on topics like derivate and integral and all other topics with their solution also so people of different country can solve the different type of ques of your country
Excellent presentation of the topics in a simple manner. Thanks with sincere regards DrRahul Rohtak Haryana India
Muy bien resueltas estas integrales
Do a vid on Euler-Mascheroni constant.
Where’s the link of sum of first n square roots ? In the description there’s sum of fist n squares...
Fady Omari Exactly what I said. I’ll try to find the link on Google on my own.
Just added, sorry.
blackpenredpen
Thanks!!
Unfortunately, there is no exact and finite formula. Only a series.
mrbus2007 I saw. There is no exact formula for the of powers of n unless the powers are natural numbers, although even then this formula is still given by an obscure summation involving the Bernoulli numbers.
Thanks that was so helpful 👌👌
Thank you!
There’s no indefinite integral that would follow regular integration rules. However I bet you could think of a nice summation to calculate the area for a floor function.
i like this!! great work
Thank you for two things:
That these integrals exist.
That with a little courage they can be solved.
Thanku sir
Your video is very supportive
Thanks
nice video :) can u try intergral of : sin(x)/sqrt(1+sin(x)*cos(x)
How would you do 2^X
Nice sir
Bruh .. !! What a wrist watch.... I've been looking at the watch at half of the time of the video
This is very interesting, but you are doing this by hand. I would like to see a proof of the indefinite integral of the floor function and its powers. Can you do that?
RJ
But can you do the same with {x}^2 and {x^2}?
The last problem can be solved geometrically very easily
So, Calc 1 wears Prada Lu?
But who wins the battle of the integrals?
For the second integral, if n goes to infinity, it seems the result goes to minus infinity. It sounds weird.
We are thinking with respect to x-axis.
We can solve more easily
Just rotate the screen and think w.r.t. y-axis you can see the last bar root(16) comprises a bigger rectangle whose area is 15×root(16) and then we are subtracting bar of area[ root(1)×1+root(2)×1+....+root(15)]
Just rotate
I‘ve a question... what’s floor function of 1.9999 period?
第二題用橫切的比較容易看
0 to a ∫ [ x ^(2) ] if [.] Is GIF ( greatest intergel function ) then answer is. ( ( a^(2) - 1)a - 1 ) -summation of r=2 to r= ( a^(2) - 1 ) ∑ √(r).
This is jee main key 😱
Are these functions even integrable in Riemann sense?
1/3 x^3
1/3*4^3
1/3*64
64/3
congrats for 200k subs
Vincent William Rodriguez thank you!!!
2E 5
Leon Ravenclaw ?
@@angelmendez-rivera351 2*10^5
Congratulations for 200K subscribers ...😎😎😎🤗🤗🤗🤗🤗🤗🤗🤗#yay.........
So please integrate tan(x) from 0 to (π/2)
You are like my 1/(1-x) teacher😇😇😇😇😇😇😇😇😇
Next do the integral involving the fractional part?
in that case, we can just change {x}=x-[x]
: )
I hope to see e^floor(x) and floor(e^x) integrals also...
How do you integrate floor(e^x) from x=0 to x=4??
@@entropy4048 Not sure, maybe finite sum (with a lot of elements). Maybe computer works to make it happen. Integrate from 0 to 1 would be only 2 elements.
The x-step positions are x(n)=ln(n) and height is n and summing these rectangles areas: sum (ln(n+1)-ln(n))*n from 1 to 53 + (4-ln(54))*54.
This series evaluation looks difficult, but its about 51.6798877368...
The series looks it might be "telescoping series", but I didn't try find it in detail.
There is actually accurate answer as:
216 - ln(230843697339241380472092742683027581083278564571807941132288000000000000),
but not sure how it is evaluated. It looks like some type of factorial (as there is a lot of zeroes at the end)... like 216-ln(54!)
As its sum or logaritms, it may turn product inside the ln(). Then it solves, I guess.
So if upper bound for integral is "n" the results would be:
AREA=n*floor(e^n)-ln((floor(e^n))!).
Maybe better question is what is:
integ {1-ceil(x)+floor(x)} dx from 0 to +Inf.
It could be 'indefinite' or something like -1/12, etc..
e^ln (i)
halfway there
interesting, compi can calculate the battle two:
*Integrate[Floor[x]^2, {x, 0, 4}];*
*Integrate[Floor[x^2], {x, 0, 4}];*
hi from sybermath
@@aashsyed1277 hi man, all good here, how are you? 😹😎
I'm having compi trouble after the OS upgrade from Stretch to Buster. today I'm trying to install old Stretch image... 😨😂
@@leecherlarry oh dear
@@leecherlarry fine!
you are simply awesome, 💞 from India
👍
恭喜，您已达到20万订阅者
今年，您将完成四十万订户的目标。
来自印度的问候
Sorry for mistakes
Shubham Gupta
Thanks.
you can just reply in English.
@@blackpenredpen I love Chinese language that's why I reply in Chinese
Dude that 4 looks that psi
I think these are easy....let describe a tough one
Aww you almost clicked on the problem where I had top solution ;( so close.
please can you solve this problem
∬⌊x+y⌋ dx dy
01 0 1
so integral = sum with floor
200k😎😎😎🥳🥳🥳🥳🥳🤠🤠🤠😇😇😇
Love from India🇮🇳🇮🇳🇮🇳🇮🇳
It isn't that hard to find a general formula...
Too easy
am i the only one who thinks that the answer should be 1 pm but the answer isn't there?
First!