I'm a high school student preparing for the JEE(Advanced) exam. It is a very famous competitive exam in India, considered as very difficult one. This video has been very helpful. Please upload the video explanations to the other problems of the integral bee if possible.
@@ArifSolvesIt waitnthe denominator is R minus 1/2 and numerator is R plus 1 if you ignore the 3/4 sonitnis only APPROXIMATELY R/R..and so this are is approximate because technically technically function blows up at x equals pi/2 the upper bound since the value of the function at that point goes to infinty sontefubicslly thus area is undefined because it is unbounded truly technically at the upper bound point of x =pi/2 .since tan pi/2 goes on finite and cube root of infinity is still infinty divided by 1 in the denominator since youbhave sine pi/2 + cosine pi/2 which reduced tonl 1 + 0...see what I mean?
Dear Arif- you went through the steps like a magician- loved it. Just a small request-the last step of 'cancelling' the infinites in integral I_2 - you have implicitly used CAUCHY's PRINCIPAL VALUE. Could you make similar videos involving improper integrals that apparently diverge, but can be indeed evaluated by the above technique and tell us when such methods to be used. In standard text Calc 2 text books the result for your integral would be infinity!
I would substitute y^3=tan(x), then i would integrate by parts with u=y and dv = 3y^2/(t^3+1)^2 dy Lastly I would use partial fraction decomposition to calculate integral Int(1/(y^3+1),y=0..infinity) This means that I would calculate this integral the same way
thanks a lot for your comment; it is a very nice substitution. In the end, the last integral is sth we all need to solve I think. I do not know if there was an easier way to get the answer by making another smart substitution!
@@ArifSolvesIt Integral Int(1/(y^3+1),y=0..infinity) can be calculated without partial fractions 1. Int(1/(y^3+1),y=0..infinity) = Int(1/(y^3+1),y=0..1)+Int(1/(y^3+1),y=1..infinity) Int(1/(y^3+1),y=1..infinity) , substitution y = 1/u 2. Int(1/(y^3+1),y=0..infinity) Substitution u = (1-y)/(1+y) After this substitution and then linearity of the integral in one of the integrals we have even function on symmetric interval around zero , in the second one we have odd function on symmetric interval around zero Finally we should get integral 2Int(1/(1+3u^2),u=0..1)
@@holyshit922 why don't you do y equals tangent x instead if y equals tan x^1/3...that's how instarted....can't you then do integration by parts with that and isn't that easier or just as manageable? Thanks for sharing.
Many thanks! It is very interesting. Recently it was published a book about MIT integration bee. The title is "MIT Integration Bee Solutions of Qualifying Tests from 2010 to 2023 ". I think it will be useful to see it
I used contour integration instead of partial fractions. There's a pole at z=-1, so the contour is one which runs along the line from z=0 to z=k and around a circle centered at the origin to z=k*zeta and then the line back to z=0, where zeta is the nontrivial cube root of 1, that is the root of z^2+z+1=0.
very good. I also have another video where I used the contour integral to solve that specific integral as well; please check: ruclips.net/video/fLkNONqsVyA/видео.html probably the contour integration leads to the answer faster.
thanks a lot, I will try to put more videos alike. I am currently working on some Physics projects but I will try to put more integration videos as well.
Thanks sir, found your channel ,feeling great Pla bring more problems As I want to tell you that I m preparing for jee advanced In which simple but hard looking problems came!!!! If you see these types of problems bring them here we all solve them with you!!!!!
At 20:47 you've used the fact that the sum of a limit is the limit of the sum, however I thought that law only applies when both individual limits exist, in this case both of the individual limits equal to infinity, so wouldn't that mean the law does not apply here?
please note that I intentionally tried to write both limits in terms of the "t" variable, and then called the upper limit to be R for the t variable. Of course, you can not say that if a certain limit is infinity and another one is also an infinity when you subtract them the result is zero. This is of course not true. But think of a function f(x)=x-x and take the limit lim(x->\infty) f(x). In this case, the limit is of course 0, since f(x) is strictly zero. Another example would be f(x)=ln(x^2+1) - ln(x^2), and you are asked to take the limit x -> \infty. In this case, you can write the function as f(x)=ln(1+1/x^2) and when you take the limit you get zero, since ln(1)=0. This is basically what I did. I found the integral in terms of the upper limit R, and then took the limit R go to \infty. What made you confused is probably to see different integrals separately. They are actually the part of just one integral, separated for the sake of being able to solve it. I just showed that when we write a single integral as the composition of different integrals, the individual integrals may lead you to infinities if you tend to solve them by disregarding the other pieces. In that sense, you can realize that we actually wrote the limit of sums (because we have only one integral to start with) as the sum of the limits when we decomposed the integral into different integrals, and then combined them back into the limit of sums.
It took me a while to see that too. That is the beauty of integration; it is not as easy as differentiating :-) sometimes it may take ages to see a substitution or a simple trick that would help you solve it in seconds.
@@ArifSolvesIt Doesn't that make yiu HATE HATE this.that takes.too.damn long to see....what if I am an impatient genius who doesn't need to be spending that much time on any one problem Ora lot of problems...how do you not get bored and fed up with it? And why not just do a subsotutituion of u equals tan x to begin with and then can't you solve u^1/3/(u+1)^2 using integration by parts..wouldn't that work? Thanks for sharing.
thanks a lot for your comment. I will try to do my best. I am currently working on some Physics lectures. But I will put more integration videos in time.
very good. there are other ways too, like the residue theorem: ruclips.net/video/fLkNONqsVyA/видео.html if you are competing, you should definitely find the fastest way possible. Unfortunately, I am getting old and not as fast as I used to be :-) so I just present different ways to those who are just curious.
great video, thanks for sharing. It proves how handy the definitions of beta and gamma functions can be. Sometimes it may be difficult to remember the relations / definitions of beta and gamma functions, that's why I just wanted to present different techniques. One other possibility is to use the residue theorem: ruclips.net/video/fLkNONqsVyA/видео.html (yet it is still not as fast as the beta function route) :-)
there are different ways to reduce it down to \int dt/(1+t^3). I couldn't come up with any other tricks which would make me end up with an integral other than \int dt/(1+t^3). The rest can be solved either using the Residue Theorem which I also demonstrated on my channel in a different video (ruclips.net/video/fLkNONqsVyA/видео.html) or using the partial fraction decomposition as I showed here. In my videos, I try to show all the steps in a more elaborative way (thus I am not trying to solve them fast) but some people are extremely fast to pass most of those steps in seconds. For example, there are those (I am definitely not one of them) who can do partial fraction decomposition in few seconds. Some are also very fast in using the residue theorem. So, in short, it is not impossible to carry out any of these calculations in less than 4 minutes.
@@ArifSolvesIt if i recall there are tricks in partial fraction decomposition, my calc teacher could do these in 30 seconds often times. He was also brilliant and worked for the federal government in missile guidance systems so he was always just a step above all the other teachers at my college. Absolutely brilliant man, so i guess with enough tricks and practice the partial decomposition or beta functions come naturally. Maybe thats how they reduce these down to less than 4 minutes! crazy feats of humans!
yes, you can solve it using the beta function. If you know the residue theorem, it is also another alternative ruclips.net/video/fLkNONqsVyA/видео.html it is good to know all possible ways, then you can choose whatever you want.
I'm a high school student preparing for the JEE(Advanced) exam. It is a very famous competitive exam in India, considered as very difficult one. This video has been very helpful. Please upload the video explanations to the other problems of the integral bee if possible.
thanks a lot for your reply. I will try to post solutions to other problems in time. all the best in your studies for the JEE exam.
@@ArifSolvesIt waitnthe denominator is R minus 1/2 and numerator is R plus 1 if you ignore the 3/4 sonitnis only APPROXIMATELY R/R..and so this are is approximate because technically technically function blows up at x equals pi/2 the upper bound since the value of the function at that point goes to infinty sontefubicslly thus area is undefined because it is unbounded truly technically at the upper bound point of x =pi/2
.since tan pi/2 goes on finite and cube root of infinity is still infinty divided by 1 in the denominator since youbhave sine pi/2 + cosine pi/2 which reduced tonl 1 + 0...see what I mean?
shut up
@AbhinavChalla : How was your result Bro❤
Dear Arif- you went through the steps like a magician- loved it.
Just a small request-the last step of 'cancelling' the infinites in integral I_2 - you have implicitly used CAUCHY's PRINCIPAL VALUE.
Could you make similar videos involving improper integrals that apparently diverge, but can be indeed evaluated by the above technique and tell us when such methods to be used.
In standard text Calc 2 text books the result for your integral would be infinity!
I would substitute y^3=tan(x),
then i would integrate by parts with u=y and dv = 3y^2/(t^3+1)^2 dy
Lastly I would use partial fraction decomposition to calculate integral
Int(1/(y^3+1),y=0..infinity)
This means that I would calculate this integral the same way
thanks a lot for your comment; it is a very nice substitution. In the end, the last integral is sth we all need to solve I think. I do not know if there was an easier way to get the answer by making another smart substitution!
@@ArifSolvesIt
Integral Int(1/(y^3+1),y=0..infinity) can be calculated without partial fractions
1.
Int(1/(y^3+1),y=0..infinity) = Int(1/(y^3+1),y=0..1)+Int(1/(y^3+1),y=1..infinity)
Int(1/(y^3+1),y=1..infinity) ,
substitution y = 1/u
2.
Int(1/(y^3+1),y=0..infinity)
Substitution u = (1-y)/(1+y)
After this substitution and then linearity of the integral in one of the integrals we have even function on symmetric interval around zero , in the second one we have odd function on symmetric interval around zero
Finally we should get integral
2Int(1/(1+3u^2),u=0..1)
@@holyshit922 why don't you do y equals tangent x instead if y equals tan x^1/3...that's how instarted....can't you then do integration by parts with that and isn't that easier or just as manageable? Thanks for sharing.
@@leif1075 I prefer to not have fractions in the power
When you substitute y=tan x you will not get rational function to integrate
@@holyshit922 right Ok but you cam still integrate thst rational power using integration by parts I think right? IF you did y=tangent x instead.
Many thanks!
It is very interesting.
Recently it was published a book about MIT integration bee. The title is "MIT Integration Bee Solutions of Qualifying Tests from 2010 to 2023 ".
I think it will be useful to see it
Thanks for sharing the information. I will check it out.
I used contour integration instead of partial fractions. There's a pole at z=-1, so the contour is one which runs along the line from z=0 to z=k and around a circle centered at the origin to z=k*zeta and then the line back to z=0, where zeta is the nontrivial cube root of 1, that is the root of z^2+z+1=0.
very good. I also have another video where I used the contour integral to solve that specific integral as well; please check: ruclips.net/video/fLkNONqsVyA/видео.html probably the contour integration leads to the answer faster.
Awesome!!
thanks a lot.
Tebrikler hocam
Güzel çözüm 👏
cok tesekkurler yorumunuz icin :-) :-) :-)
Sir,i use the formulas containing beta and gamma functions in: x is (0,inf.)(1/(1+x^3))
this sounds like a very good strategy. I will look into that. Thanks a lot for your suggestion.
Love the video, please more MIT integration Bee problems
thanks a lot, I will try to put more videos alike. I am currently working on some Physics projects but I will try to put more integration videos as well.
Thanks sir, found your channel ,feeling great
Pla bring more problems
As I want to tell you that I m preparing for jee advanced
In which simple but hard looking problems came!!!!
If you see these types of problems bring them here we all solve them with you!!!!!
as time permits, I will try to solve more problems.
At 20:47 you've used the fact that the sum of a limit is the limit of the sum, however I thought that law only applies when both individual limits exist, in this case both of the individual limits equal to infinity, so wouldn't that mean the law does not apply here?
please note that I intentionally tried to write both limits in terms of the "t" variable, and then called the upper limit to be R for the t variable. Of course, you can not say that if a certain limit is infinity and another one is also an infinity when you subtract them the result is zero. This is of course not true. But think of a function f(x)=x-x and take the limit lim(x->\infty) f(x). In this case, the limit is of course 0, since f(x) is strictly zero. Another example would be f(x)=ln(x^2+1) - ln(x^2), and you are asked to take the limit x -> \infty. In this case, you can write the function as f(x)=ln(1+1/x^2) and when you take the limit you get zero, since ln(1)=0. This is basically what I did. I found the integral in terms of the upper limit R, and then took the limit R go to \infty. What made you confused is probably to see different integrals separately. They are actually the part of just one integral, separated for the sake of being able to solve it. I just showed that when we write a single integral as the composition of different integrals, the individual integrals may lead you to infinities if you tend to solve them by disregarding the other pieces. In that sense, you can realize that we actually wrote the limit of sums (because we have only one integral to start with) as the sum of the limits when we decomposed the integral into different integrals, and then combined them back into the limit of sums.
Ooh, sec^2 is a good move. I fell into the trap of expanding the bottom to cos^2(x)+2sin(x)cos(x)+sin^2(x)=1+sin(2x)
It took me a while to see that too. That is the beauty of integration; it is not as easy as differentiating :-) sometimes it may take ages to see a substitution or a simple trick that would help you solve it in seconds.
@@ArifSolvesIt Doesn't that make yiu HATE HATE this.that takes.too.damn long to see....what if I am an impatient genius who doesn't need to be spending that much time on any one problem Ora lot of problems...how do you not get bored and fed up with it? And why not just do a subsotutituion of u equals tan x to begin with and then can't you solve u^1/3/(u+1)^2 using integration by parts..wouldn't that work? Thanks for sharing.
Thanks this is youseful for my a level exam soon :D
you're welcome. thanks a lot for your comment.
I am a student of college stage .I am taking preparation for BUET admission.please upload all MIT intregal bee problem if possible sir.
thanks a lot for your comment. I will try to do my best. I am currently working on some Physics lectures. But I will put more integration videos in time.
just let tanx=t ,then the integral is Beta(4/3 ,2/3)=1/3 *pi*csc(pi/3) ,do it in my mind . 3mins is enough.
very good. there are other ways too, like the residue theorem: ruclips.net/video/fLkNONqsVyA/видео.html if you are competing, you should definitely find the fastest way possible. Unfortunately, I am getting old and not as fast as I used to be :-) so I just present different ways to those who are just curious.
Very nice solution sir, I’ll admit this integral first stumped me 😅
thanks a lot for your comment. It took me a while too. In short, I need to admit it took me way more than 4 minutes to solve it the first time :-)
Answer is 2 sqrt(3) × pi / 9
Solve it in 4 mins. bro. It's easy to solve it in 30 mins.:))
Its alot easier to solve this using the beta function:
ruclips.net/video/J20crVcWra0/видео.html
great video, thanks for sharing. It proves how handy the definitions of beta and gamma functions can be. Sometimes it may be difficult to remember the relations / definitions of beta and gamma functions, that's why I just wanted to present different techniques. One other possibility is to use the residue theorem: ruclips.net/video/fLkNONqsVyA/видео.html (yet it is still not as fast as the beta function route) :-)
@@ArifSolvesIt all the same I enjoyed your brute force treatment of the integral. Refreshing to see some old school integration.
There must be an easy trick to do that integral, because your solution is impossible to do in 4 min, not even in 10.
there are different ways to reduce it down to \int dt/(1+t^3). I couldn't come up with any other tricks which would make me end up with an integral other than \int dt/(1+t^3). The rest can be solved either using the Residue Theorem which I also demonstrated on my channel in a different video (ruclips.net/video/fLkNONqsVyA/видео.html) or using the partial fraction decomposition as I showed here. In my videos, I try to show all the steps in a more elaborative way (thus I am not trying to solve them fast) but some people are extremely fast to pass most of those steps in seconds. For example, there are those (I am definitely not one of them) who can do partial fraction decomposition in few seconds. Some are also very fast in using the residue theorem. So, in short, it is not impossible to carry out any of these calculations in less than 4 minutes.
@@ArifSolvesIt thanks, ok, not impossible but extremely hard.
@@ArifSolvesIt if i recall there are tricks in partial fraction decomposition, my calc teacher could do these in 30 seconds often times. He was also brilliant and worked for the federal government in missile guidance systems so he was always just a step above all the other teachers at my college. Absolutely brilliant man, so i guess with enough tricks and practice the partial decomposition or beta functions come naturally. Maybe thats how they reduce these down to less than 4 minutes! crazy feats of humans!
man it is fking beta func
yes, you can solve it using the beta function. If you know the residue theorem, it is also another alternative ruclips.net/video/fLkNONqsVyA/видео.html it is good to know all possible ways, then you can choose whatever you want.