I wish someone could do something like this online, where anyone can qualify and then be part of this, dont know how many people would be interested though
I didn't like quite a lot, the problems of this year are much more technical and you need to spend more time (for example) to understand the graph of function. Such problems need min 4 min for solving.
@@МаркТурцынский you’re right, I was only able to solve 7 of the problems in under 2 min, but bringing this mainstream in my opinion would attract a lot of people to calculus
@@jonahgrigoryan You can compare with the examples from previous competitions. They previously made much more focus on the ideas of integration not on some supplementary things like graphs. This is the overcomplexity from my point of view. May be the authors were not the same. May be they should give different time to different examples.
I think it could be interesting to see what would happen if the allotted time was double of what it is. It would make the competition less based on a lucky first approach, while still making sure the proper winners are those who can integrate fast
Man, in the two minutes available these are brutal. None are impossible to do, but most involve lots of fiddly bits you have to keep track of. (Cue all of the 'ha, in my country we do these before we can walk' comments.)
I can do most of these, but only a few I could do in the allotted time. Number 13 was a tricky one. Although I don't think the integrand can be stated in closed form, its inverse can on that interval. Integrate along the y-axis instead of the x-axis.
I would had loved to participate, in event like this back at my university days, if we had one at it. But we may only have 5 participates, it is only in university's like MIT were the quality of teaching and quality of students is so high.
Recently it was published a book about MIT integration bee, under the title " MIT Integration Bee, Solutions of Qualifying Tests from 2010 to 2023" You can simply find it!
@@12ha Dumb thinking. Chirag is just a topper. While Newton was one of the greatest minds throughout human history. Being a topper is not a requisite for doing something extraordinary and impossible in life.
problem 14 int (floor x)x(ceiling x) from 0 to 100 (int 0*1*x dx from 0 to 1)+(int 1*2x dx from 1 to 2)+(int 2*3x dx from 2 to 3)+.....+(int (n-1)*n xdx from n-1 to n) where n=100. I=sum (n-1)n(n^2-(n-1)^2)*(1/2) from 1 to 100 (1/2)sum (n-1)n(2n-1) from 1 to 100 (1/2)*(sum 2*n^3-3n^2+n) from 1 to 100 (1/2)*(2*((n-1)n/2)^2-3*n(n+1)(2n+1)/6+n(n+1)/2))=n(n+1)/4(n(n+1)-(2n+1)+1)=(1/4)*n(n+1)(n^2-n)=(1/4)*n^2(n+1)(n-1)=1/4*n^2(n^2-1)=(n^4-n^2)/4=(100^4-100^2)/4
@@M1551NGN0 Indian people talk so much shit and there is nothing to back it up. Yeah these problems should be doable for anybody who has taken calculus, no kidding. The time pressure is what makes these hard.
@@pendragon7600 and we've had proofs for everything until your European Bros declared Axiomatic proofs more important than Empirical proofs. And anyways i was referring to the inconsistency of the quote that the original person of this comment wrote by telling them politely that this is not exactly something that only MIT minds could do.
draw or imagine the two functions being compared, then integrate one for x where it is greater (or smaller), and integrate the other for the rest of the x in the limits of integration. For the min part of Problem 3, integrate 2x from 0 until 1, then the other fraction gets smaller than 2x so integrate the fraction from 1 to 3. add the two togetter for the total integral of the min funciton. or conceptually, find the area under the highest (or lowest) of the two lines for any part of the graph
max( function1, function2 ) means you take, precisely, the maximum value between both functions in a determined interval. For example, consider the function max(x, 2-x). If you see the graph of both functions, or just make x=2-x, we can see they intersect in P(1,1). See that before x=1 (interval (-inf,1)) 2-x is bigger than x, so the function takes the values of 2-x. Same way, see that after x=1 (interval (1,inf)) x is bigger than 2-x, so the function takes the values of x. It's easier to think which function goes "over" the other on the graph. Now, for example, imagine you try to integrate from 0 to 2. So what we have to do is integrate from 0 to 1 the function 2-x (since it's bigger than "x") and sum that to the integral from 1 to 2 of x (since it's bigger than 2-x). Hope this helps 👍
max( function1, function2 ) means you take, precisely, the maximum value between both functions in a determined interval. For example, consider the function max(x, 2-x). If you see the graph of both functions, or just make x=2-x, we can see they intersect in P(1,1). See that before x=1 (interval (-inf,1)) 2-x is bigger than x, so the function takes the values of 2-x. Same way, see that after x=1 (interval (1,inf)) x is bigger than 2-x, so the function takes the values of x. It's easier to think which function goes "over" the other on the graph. Now, for example, imagine you try to integrate from 0 to 2. So what we have to do is integrate from 0 to 1 the function 2-x (since it's bigger than "x") and sum that to the integral from 1 to 2 of x (since it's bigger than 2-x). Hope this helps 👍
Problem 13 int sin(x-sin(x-sin(x- ...)))dx from 0 to pi/2+1. sin(x-sin(x-sin(x- ...)))=u sin(x-u)=u. x-u=arc sin(u) x=arc sin(u)+u dx=(1/((1-u^2)^(1/2))+1) if x=0 then u=0 if x= pi/2+1 then u=1 I=int u(1+1/((1-u^2)^(1/2))dx from 0 to 1 I=(u^2)/2 +int u/((1-u^2)^(1/2))dx=(u^2)/2-(1-u^2)^(1/2) from 0 to 1 I=1/2-(-1)=3/2
How to do Q4. I mean What I was doing was assuming it will continue till infinity so x=1-1/x etc. I want to know 1) why cant we extend it to infinity 2) how to solve it
Right next to him the guy on the second board Luke robataille 4 time international maths olympiad gold medalist Also the winner of this years integration bee
@@piyushsharma8442He did not reach finals ironic how literally all Indians on the comment section keep saying how easy it is yet their best could not solve it.If India is so great stay there stop coming to the US
@@garvrajput9983 They are easy question but the people solving them are god level students so I think yeah we can solve but surely not as fast as them
I wish someone could do something like this online, where anyone can qualify and then be part of this, dont know how many people would be interested though
Yes, that would be a nice idea, Johan.
I didn't like quite a lot, the problems of this year are much more technical and you need to spend more time (for example) to understand the graph of function. Such problems need min 4 min for solving.
@@МаркТурцынский you’re right, I was only able to solve 7 of the problems in under 2 min, but bringing this mainstream in my opinion would attract a lot of people to calculus
@@jonahgrigoryan You can compare with the examples from previous competitions. They previously made much more focus on the ideas of integration not on some supplementary things like graphs. This is the overcomplexity from my point of view. May be the authors were not the same. May be they should give different time to different examples.
I would be interested, cue me if you come up with something like this
I think it could be interesting to see what would happen if the allotted time was double of what it is. It would make the competition less based on a lucky first approach, while still making sure the proper winners are those who can integrate fast
2 minute is very fast to solve integrals. I think it should be atleast 5-10 minutes per problem.
Man, in the two minutes available these are brutal. None are impossible to do, but most involve lots of fiddly bits you have to keep track of. (Cue all of the 'ha, in my country we do these before we can walk' comments.)
These are not hard though
It is just the time that makes it hard, if it is 3-5 min, it is gonna be so easy
in my country we do these before we can walk
In America, we clearly can do these before we walk
@@maxgeorge1463 😢
I enjoy watching stuff like this too much, it's so good
I can do most of these, but only a few I could do in the allotted time. Number 13 was a tricky one. Although I don't think the integrand can be stated in closed form, its inverse can on that interval. Integrate along the y-axis instead of the x-axis.
Having seen the finals, so strange to see the future winner struggling and slowly getting mad
56:13 that was so wholesome and the guy took it like a champ hahaha
I feel so much better about my math knowledge after watching this. 😂
this takes me back when my friend won the 2006 mit integration bee !!!!
@aiien8768
Neat! How long did you know him?
I would had loved to participate, in event like this back at my university days, if we had one at it. But we may only have 5 participates, it is only in university's like MIT were the quality of teaching and quality of students is so high.
Recently it was published a book about MIT integration bee, under the title " MIT Integration Bee, Solutions of Qualifying Tests from 2010 to 2023"
You can simply find it!
Can you send me the link?
@@danielespinosa869 I think it is forbidden to send links in the comments. But by the title you can simply find it
Anyone Noticed Chirag Falor
He is the Air 1 For jee Advance 2020
And he is not the Grand Integrator too!
@@u.v.s.5583 woh toh newton h india ka
Yes
@@12ha Dumb thinking. Chirag is just a topper. While Newton was one of the greatest minds throughout human history. Being a topper is not a requisite for doing something extraordinary and impossible in life.
WHAT?! *WHAT?!?*
Must continue Posting ...🙏🙏 Incredible
11:18 lol she’s snapping it to her friends
In 2 min it's really hard to do these especially when you are not trained i hope that they practice beforehand
problem 14 int (floor x)x(ceiling x) from 0 to 100 (int 0*1*x dx from 0 to 1)+(int 1*2x dx from 1 to 2)+(int 2*3x dx from 2 to 3)+.....+(int (n-1)*n xdx from n-1 to n) where n=100.
I=sum (n-1)n(n^2-(n-1)^2)*(1/2) from 1 to 100 (1/2)sum (n-1)n(2n-1) from 1 to 100 (1/2)*(sum 2*n^3-3n^2+n) from 1 to 100 (1/2)*(2*((n-1)n/2)^2-3*n(n+1)(2n+1)/6+n(n+1)/2))=n(n+1)/4(n(n+1)-(2n+1)+1)=(1/4)*n(n+1)(n^2-n)=(1/4)*n^2(n+1)(n-1)=1/4*n^2(n^2-1)=(n^4-n^2)/4=(100^4-100^2)/4
Sure, because everyone knows sum(n^3) of the top of their head...
This should be an Olympic sport
"Behind every great unsolved integral there is a MIT mind!"
L take
XD
Nah homie problem 5 is in our textbooks in Indian schools
@@M1551NGN0 Indian people talk so much shit and there is nothing to back it up. Yeah these problems should be doable for anybody who has taken calculus, no kidding. The time pressure is what makes these hard.
@@pendragon7600 and we've had proofs for everything until your European Bros declared Axiomatic proofs more important than Empirical proofs. And anyways i was referring to the inconsistency of the quote that the original person of this comment wrote by telling them politely that this is not exactly something that only MIT minds could do.
What if they had the Triple Integration Bee?
It’s a good thing differential calculus is more important
Great job 👍
This could be more like a challenge I guess or maybe an integration challenge in this class
Loved it .
Some of these questions are really easy for JEE aspirants.
17:38
Pointing at each contestant: Wrong.. Wrong.. Wrong.. Wrong 😂
These bee chess stole my icon!
Anyone noticed anton trygub? He's a LGM on codeforces
Sorry for my ignorance, but how in the world do u integrate a min or max function?
draw or imagine the two functions being compared, then integrate one for x where it is greater (or smaller), and integrate the other for the rest of the x in the limits of integration. For the min part of Problem 3, integrate 2x from 0 until 1, then the other fraction gets smaller than 2x so integrate the fraction from 1 to 3. add the two togetter for the total integral of the min funciton.
or conceptually, find the area under the highest (or lowest) of the two lines for any part of the graph
max( function1, function2 ) means you take, precisely, the maximum value between both functions in a determined interval. For example, consider the function max(x, 2-x). If you see the graph of both functions, or just make x=2-x, we can see they intersect in P(1,1). See that before x=1 (interval (-inf,1)) 2-x is bigger than x, so the function takes the values of 2-x. Same way, see that after x=1 (interval (1,inf)) x is bigger than 2-x, so the function takes the values of x. It's easier to think which function goes "over" the other on the graph. Now, for example, imagine you try to integrate from 0 to 2. So what we have to do is integrate from 0 to 1 the function 2-x (since it's bigger than "x") and sum that to the integral from 1 to 2 of x (since it's bigger than 2-x).
Hope this helps 👍
Chinese are leading the MiT as well.Good job guys.
What about Indians
i didn't understand the first one ..!! what does "max(sinx,cox)" mean?
max( function1, function2 ) means you take, precisely, the maximum value between both functions in a determined interval. For example, consider the function max(x, 2-x). If you see the graph of both functions, or just make x=2-x, we can see they intersect in P(1,1). See that before x=1 (interval (-inf,1)) 2-x is bigger than x, so the function takes the values of 2-x. Same way, see that after x=1 (interval (1,inf)) x is bigger than 2-x, so the function takes the values of x. It's easier to think which function goes "over" the other on the graph. Now, for example, imagine you try to integrate from 0 to 2. So what we have to do is integrate from 0 to 1 the function 2-x (since it's bigger than "x") and sum that to the integral from 1 to 2 of x (since it's bigger than 2-x).
Hope this helps 👍
1:08:58
This one was easy why Noone could solve it?
Problem 13
int sin(x-sin(x-sin(x- ...)))dx from 0 to pi/2+1. sin(x-sin(x-sin(x- ...)))=u
sin(x-u)=u. x-u=arc sin(u) x=arc sin(u)+u dx=(1/((1-u^2)^(1/2))+1) if x=0 then u=0 if x= pi/2+1 then u=1 I=int u(1+1/((1-u^2)^(1/2))dx from 0 to 1 I=(u^2)/2 +int u/((1-u^2)^(1/2))dx=(u^2)/2-(1-u^2)^(1/2) from 0 to 1 I=1/2-(-1)=3/2
????
8:53 who else saw 16)Chirag Falor
i could do the first one in 10 minutes.
How to do Q4. I mean What I was doing was assuming it will continue till infinity so x=1-1/x etc. I want to know 1) why cant we extend it to infinity 2) how to solve it
Or try cengage problem it will help you to crack integration
@UA-camOfficialHandle x - LNX
@@BariScienceLab❤
18:05 😂😂😂
What if I added a C constant, and my answer is right. Will they take my answe?
33:30 bro what is that
That (n r) represents nCr=n!/r!(n-r)!
@@satishchaudhary7978 no like the problem there is no line in the fraction
what is max(sin(x),sin(x))
Cant we just flip the denominator in 2nd ques and substitute the limits directly? 13:17
Also how to solve q18
In question 16 board no. 1 chirag falor
jee advanced air 1
Red shirt?
@@Sa.m1498 yes
ok ? no one asked tho
@@pushkarmishra3386 no one say to you to comment here.
Right next to him the guy on the second board Luke robataille
4 time international maths olympiad gold medalist
Also the winner of this years integration bee
How do i lesrn this if i only know a bit of pre calculus?
Start with the basics of calc I and build onto it with calc II
bruhhhh luke 👺👹
When u are a JEE Advanced Aspirant🔥🔥🔥🔥🔥🔥🔥🔥
AIR 1 COULDNT GET SELECTED SOMETHING FALOR OVERFANSTSY COUNTRY WITH OVERHYPRD EXAM
@@Won49he reached finals you better think about yourself before putting a comment on a world class exam
@@piyushsharma8442He did not reach finals ironic how literally all Indians on the comment section keep saying how easy it is yet their best could not solve it.If India is so great stay there stop coming to the US
You have money. No nu no petrol price don't try
OK and what is this good for?
got the second in about 3 mins
thisssuuhhh math is rather easy i coud do it in less than 5 second
Chirag up there
Presenter talks way too much. Over 9 minutes before first integral is shown!
it is easy for an jee aspirants who solving reference books like cengage etc
yes thats why AIR 1 JEE-A 2020 didnt won. right?
@@garvrajput9983 the single student cant solve it not means that the level of question is easy although it can be solve by me also
@@garvrajput9983 They are easy question but the people solving them are god level students so I think yeah we can solve but surely not as fast as them
Jee advanced aspirant? 😅
CHIRAG FALOR IIT JEE 2014 ALL INDIA RANK 1🔥🔥