Only a teacher who loves his subject can make more teachers. You are awesome sir. Love you from India

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!

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.

Awesomely done!!!

That's a 600k+ subscribers ! Congratulations. I would have drawn the parabola y=x^2 for the second integral.

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

Excellent presentation of the topics in a simple manner. Thanks with sincere regards DrRahul Rohtak Haryana India

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

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

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).

Thanks that was so helpful 👌👌

i like this!! great work

Will you ever be doing a video on Lagrange's multipliers?

Thank you!

Thanku sir Your video is very supportive Thanks

Nice sir

Muy bien resueltas estas integrales

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

Bruh .. !! What a wrist watch.... I've been looking at the watch at half of the time of the video

Where’s the link of sum of first n square roots ? In the description there’s sum of fist n squares...

Thank you for two things: That these integrals exist. That with a little courage they can be solved.

The last problem can be solved geometrically very easily

congrats for 200k subs

Do a vid on Euler-Mascheroni constant.

you are simply awesome, 💞 from India

Congratulations for 200K subscribers ...😎😎😎🤗🤗🤗🤗🤗🤗🤗🤗#yay......... So please integrate tan(x) from 0 to (π/2) You are like my 1/(1-x) teacher😇😇😇😇😇😇😇😇😇

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.

nice video :) can u try intergral of : sin(x)/sqrt(1+sin(x)*cos(x)

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

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

2E 5

1/3 x^3 1/3*4^3 1/3*64 64/3

Next do the integral involving the fractional part?

halfway there

How would you do 2^X

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).

But who wins the battle of the integrals?

interesting, compi can calculate the battle two: *Integrate[Floor[x]^2, {x, 0, 4}];* *Integrate[Floor[x^2], {x, 0, 4}];*

But can you do the same with {x}^2 and {x^2}?

I hope to see e^floor(x) and floor(e^x) integrals also...

👍

For the second integral, if n goes to infinity, it seems the result goes to minus infinity. It sounds weird.

So, Calc 1 wears Prada Lu?

I‘ve a question... what’s floor function of 1.9999 period?

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

e^ln (i)

Dude that 4 looks that psi

Aww you almost clicked on the problem where I had top solution ;( so close.

I think these are easy....let describe a tough one

200k😎😎😎🥳🥳🥳🥳🥳🤠🤠🤠😇😇😇

第二題用橫切的比較容易看

Are these functions even integrable in Riemann sense?

so integral = sum with floor

Love from India🇮🇳🇮🇳🇮🇳🇮🇳

please can you solve this problem ∬⌊x+y⌋ dx dy 01 0 1

am i the only one who thinks that the answer should be 1 pm but the answer isn't there?

First!

Too easy

It isn't that hard to find a general formula...