I remember in high school, I used to look at problems like this, my eyes would get wide, and I'd think to myself "damn, that math is probably super far away for me, I wish I could know this stuff" Well, it's not really that big of a deal, but finally being able to understand this shit makes me happy for some reason...
It makes you feel comfortable because a type 5 likes to gather information to feel comfortable. This is just a model and not fully accurate, but I'm seeing connections with a key trait of your personality that would be a type 5
Thank you, Alex! I had to decide between including an intuitive explanation on polar transformation (and using up a lot of time) or just resorting to "trust me, it's this." I opted for the former--after all, a quality in intuition always outweighs a shorter watch time.
@@LetsSolveMathProblems ya, though very non-trivial and requires lots of complicated maths to prove properly, the explanation is good enough to compensate for the lack of triviality. I just finished calc 2 and on my way to 3 btw.
There is a easier way to do this. If you have learned statistics, then you should know that the area under the normal distribution curve must be 1. If we tried to do completing square of the exponent to form it like the equation of normal curve, we can easily found the equation is similar to the normal curve with mean = -0.8 and standard deviation = 0.5. And the rest of calculation is just very basic. I solved the problem in this way within 5 minutes, instead of going through multi-variable cal procedure
You are absolutely right. It is not very hard to manipulate the equation so that we have a constant times the integral of pdf of the normal curve, from which the solution is immediate. However, I do wish to note that the proof of the pdf of normal curve having the necessary conditions of pdf relies on a method similar to the one in the video (it is not hard to do so using the substitution to z-scores); that is, the method using the normal pdf still does everything in the video because the primary information--the fact that area under normal pdf is one--can be seen as the result of the multivariable calculus procedure. Nevertheless, it is an excellent shortcut to use, especially if you are pressed for time. Thank you for commenting the method!
The area of the normal distribution curve being 1 is a consequence of the solution to the Gaussian integral, not the other way around. In fact, the normal distribution IS a Gaussian curve, with a normalization constant. That's why it's called the Gaussian distribution.
when you were at integral e^(-2x²) you could use u=sqrt(2)x and it would turn into the literal gaussian integral of which you already know is sqrt(pi). Of course for the substitution you'd divide by sqrt(2) to account for the chain rule, so you'd end up with sqrt(pi)/sqrt(2) = sqrt(pi/2)
i had been messing with this integral for the whole damn day and all i had to do is complete the damn square -_- the rest follows up quite easily. Thanks for this, im pretty hooked on mit integration bee integrals.
You can also do a u-sub : u = sqrt2 *(x+5/4) (it comes from the canonical form of -x²-5x-3). The bounds of the integral do not change and it gives the Gaussian integral again 🫦
Excellent explanation on evaluating the Gaussian integral. I like the visual aids you used to highlight the intuitive discussion and the three comparisons made between Cartesian and Polar coordinate systems. You did a great job! Thank you.
I m an IIT JEE aspirant. So We have only learnt simple integrals only. But you have explained it very good. I was having curious of integrating e^-x^2.
At 4:00: x and y are not just some random variables, they need to be Cartesian coordinates for the spherical coordinates to work. That puts limitations on the Gaussian integral because it would not work in singularities.
I love these videos, your explanations are very clear and thorough even for people who haven't seen this type of integral before but for those who have It doesn't take long to solve If you know the Gaussian Integral well, 1)Complete the Square 2) pull out the constant 3) Apply u substitution and you're done. Took me about a minute.
Thanks for the video, it really helped me understand polar coordinates (something I never actually understood). I wouldn't have had a clue on how to do this kind of problem before. Your solution was really nice, but after watching it i noticed that you could've made it simpler after you got to the integral of e^(-2x^2) dx by using u-substitution. If u^2=2x^2 then 2x*dx=u*du, dx=u/(2x)*du, dx=1/sqrt2*du. This makes 1/sqrt2 * int e^-u^2 du which gives sqrt(pi/2)
You are doing wonderful job, even though I am not a science or maths student but I like watching your videos ,please make videos on probability , permutations and combinations.
I think it should be clarified why the equality @5:45 is allowed. It normally isn't allowed for the same reason you can't multiply different summations together. However, in this case, because the outcome of the integral in question converges, ie the outcome is a constant, and because you can pull constants in and out of integrals, you can simplify it into one integrand as long as the variables being integrated over are different. There's a pretty good explanation here: math.stackexchange.com/questions/549923/how-the-product-of-two-integrals-is-iterated-integral-int-cdot-int-iint
Sorry, I'm not really familiar with "MIT integration bee", but how long do you have time to solve it during the competition? I mean, solving this in maybe 3 minutes or so, would be absolutely insane.
Kosekans Well, if you consider that he "wasted" 15 minutes deriving the Gauss Integral (he even explained polar coords and how to evaluate the Jacobian) and some useless steps afterwards that he did to make things look easier and more evident, this integral can actually be evaluated in 2 minutes, maybe less:) Needless to say, the Gauss Integral is very famous and extremely important in statistics since high school
I'm in Grade 11 and I really want to learn things like this but we are limited to the curriculum can you maybe consider creating "math for dummies" videos pleaaaaase I really do enjoy maths
Claudio Gomes I'm just in 7 grade and I also want to learn things like these. Though the curriculum is limited for a std. Who cares you do what you want. And your idea is great dude
You can do it with only substitutions knowing the antiderivative of an exponential and the answer to the Gaussian integral oh yeah and some algebraic manipulation
At the end, why didnt you say, u = sqrt(2)*(x+4/5)? Then you could use what you found out before becuse you would have e^(-u^2) And you would end up with the same solution
The method in this video doesn't work for that. You would need the integrand to be of the form e^(-ax^2 - bx - c), for which the integral, using the method in the video, evaluates to sqrt(pi/a)*e^(1/4a)(b^2 - 4ac).
Radio TV Using the same method as in the video. So, given your integral is of the form int(e^(-ax^2-bx-c)), you do the following: Factor out e^(-c) since it's constant = e^(-c) * int(e^(-ax^2-bx))dx Now rewrite quadratic by completing the square, = e^(-c) * int(e^( -a( x + b/(2a) )^2 + (b^2/4a) ))dx Now factor out e^(b^2/4a) since it's also a constant and simplify with e^(-c) from previous step, = e^[ (1/4a)(b^2-4ac) ] * int(e^[-a( x + b/(2a) )^2)])dx Now setting I = int(e^[-a(x + b/(2a))])dx = int(e^[-a(y + b/(2a))])dy, and keeping in mind both x and y go from -inf to +inf, and simplifying resulting expression, I^2 = int(int(e^[-a( (x + b/(2a))^2 + (y + b/(2a))^2 )))]dxdy Now, to make the substituion of r^2 = x^2 + y^2, so we can convert this definite integral to polar, let's the make the substition u = x + b/(2a) and v = y + b/(2a), which now makes the definite integral, = int(int(e^[-a( u^2 + v^2 )])dxdy And now we make the substition r^2 = u^2 + v^2, and dudv = rdrdt (I'm using t here for theta) as shown in the video, and keeping in mind the bounds of the integral are now 0 to 2pi for t and 0 to inf for r, and so, = int(int((e^[-ar^2])r))drdt And now using substition (kind of running out of letters lol) k = -ar^2, dk = -2ardr => (-1/2a)dk = dr, and since the bounds of r were from 0 to inf, the bounds of k (since we set k = -ar^2) will be from 0 to -inf, and so the integral is now, = int(int((e^[k])(-1/2a)dk)dt) Now rewrite the double integral so it's a little more clear, = (-1/2a) * int(dt) * int((e^k)dk) Solving, remembering the bounds for k! = (-1/2a) * 2pi * (-1) = pi/a And now, remember that this was I^2, so I^2 = pi/a => I = sqrt(pi/a) This, btw, is the solution to the general form of the Gaussian integral int(e^[-a(x + b)^2])dx with b instead being b/(2a) which doesn't affect the result. Now that we have I = sqrt(pi/a), the original integral was e^[ (1/4a)(b^2-4ac) ] * int(e^[-a( x + b/(2a) )^2)])dx, and we just found the write part of the product, which is I, so the solution is, e^[ (1/4a)(b^2-4ac) ] * sqrt(pi/a)
Of course! I will try to look into 16 and 18 in my free time in the near future, but I cannot promise that there will be videos on them. Thank you for your recommendations!
Hey, pretty cool that Gaussian integral :DD But what makes you feel so sure it's I = sqrt(pi) and not I = - sqrt(pi) 🤔 That made me a little confused, but everything else was just perfect, congratz for being really didatic, you'll go far that way xD
Oh come on, I thought you would’ve given us more of a challenge! I could solve this easy if I just u subbed x with -(4x+5)/2/√(2) (a simple translation and dialation), leaving behind a constant times e^(-u²), which can be easily integrated using the error function, which is solvable for +-∞. Realistically, the hardest part of this would be figuring out what to substitute u (which isn’t that hard anyway if you remember -b/(2a))
You're confusing two different things. At that step, he's not making a substitution based on a relationship between x and u. He's saying the same thing that he did at 3:29, that the variable is an arbitrary dummy variable and the value of the integral is the same regardless of what we call it. i.e. It could be integral of e^-2(star)^2 d(star) instead of u and du (or x and dx) without making any difference.
I am planning on doing so tomorrow as long as my schedule stays open! I worked on the question as soon as I saw your comment, and I believe I have an elegant solution. =)
As stated in the video integrating over dxdy is best thought of as "adding up tiny rectangles". The same concept applies when we integrate in polar where you can think of adding tiny arc sections which might look like curved rectangles where the "width" of the rectangle can be easily thought of as dr, the base would not be simply dtheta obviously as that would mean an angle representing a length but instead is thought of as rdtheta where theta is in radians due to the definiton of arclengths (where pi came from ex: a circle of diameter 1 has a circumference of pi). If you were to instead use root2*r*dr*dtheta you would need to divide dtheta by root2 in order to get the right arclength (i know dividing dtheta doesnt make much sense, using a sensible angle such as pi/2 is easier to think of). Essentially it is because we use radians, as i said before using another angle system being the angle in radians divided by root 2 would make this work out
So his definite integral would end up being from 0 to 2pi/root2. Super long reply but i hope it kinda clears things up, if not maybe find a good review video on arc lengths and the unit circle. Unit circle helps with all kinds of intuitions so its worth reviewing anyways.
u need to ask viewers to subscribe to the chanel. as per what i can see your videos have like 3/4k views but u only have 1k subs. that does not make sense. u are very good as math. now get better at selling.
Your comment is one of my favorite comments of all time! You are absolutely right--My abilities as a salesperson have never been outstanding. I will consider doing more advertisements in the future. Thank you for your encouragement and suggestion! I appreciate it. =)
Wow that's unnecessary, I think it's pretty understandable albeit the awkward pronunciation, and in no way does it hinder the video's main point in any manner. I sincerely hope the teachers you've had in your life didn't find you as much of a dotard as you seem to be
![ลองสุ่มไอโฟน 990 โกงมั้ย? [ โกงมั้ยครับ ep.101 ] | DOM](http://i.ytimg.com/vi/bfsWe0j492k/mqdefault.jpg)
I remember in high school, I used to look at problems like this, my eyes would get wide, and I'd think to myself "damn, that math is probably super far away for me, I wish I could know this stuff"
Well, it's not really that big of a deal, but finally being able to understand this shit makes me happy for some reason...
Im early into high school and am thinking the exact same thing... and hopefully will have the same reaction haha
How old are u now?
A type 5
It makes you feel comfortable because a type 5 likes to gather information to feel comfortable. This is just a model and not fully accurate, but I'm seeing connections with a key trait of your personality that would be a type 5
Me too.I'm 60yrs old back to school.
At 19:43 you could've just done another u-substitution with v=sqrt(2)*u and then you would've been able to directly substitute the Gaussian integral.
he could do completing square without fraction out (-2),just (-1), an then he had not to do other v -substitution.
@@embedded_ completing the square would be harder with 2 as the coefficient of x^2
@@oximas no
e^(-2[x²-2x(5/4)+25/16-25/16]-3)
= e^(-2(x-5/4)²+25/8-3)
= e^(⅛) e^((√2(x-5/4))²)
dx = d(x-5/4) = 1/√2 d(√2(x-5/4))
e^(⅛) ₋₀₀∫⁰⁰ e^((√2(x-5/4))²)/√2 d(√2(x-5/4))
= e^(⅛)/√2 (√π)
= e^(⅛) √(π/2)
You are like Khan Academy on steroids.
hahahaha
I love this version of the gaussian integral, because the part dxdy=rdrdθ is with put using the jacobian, i love it :3
Great explanation on the transformation to polar coordinates!
Thank you, Alex! I had to decide between including an intuitive explanation on polar transformation (and using up a lot of time) or just resorting to "trust me, it's this." I opted for the former--after all, a quality in intuition always outweighs a shorter watch time.
LetsSolveMathProblems definitely!
I watched a couple of the explanation on TH-cam and his explaination is the best one so far!
@@LetsSolveMathProblems ya, though very non-trivial and requires lots of complicated maths to prove properly, the explanation is good enough to compensate for the lack of triviality.
I just finished calc 2 and on my way to 3 btw.
That integral form is common in Quantum Physics.
Integral in that form is:
(-inf→+inf)￡e^[-(ax^2 +bx)] dx
= sqrt[pi/a] * e^[(b^2)/4a)
Jabami Yumeko lol using pound for integral
@@Andrew-ri5vs lol
One of the first questions from Griffiths QM is this integral
There is a easier way to do this. If you have learned statistics, then you should know that the area under the normal distribution curve must be 1. If we tried to do completing square of the exponent to form it like the equation of normal curve, we can easily found the equation is similar to the normal curve with mean = -0.8 and standard deviation = 0.5. And the rest of calculation is just very basic.
I solved the problem in this way within 5 minutes, instead of going through multi-variable cal procedure
You are absolutely right. It is not very hard to manipulate the equation so that we have a constant times the integral of pdf of the normal curve, from which the solution is immediate. However, I do wish to note that the proof of the pdf of normal curve having the necessary conditions of pdf relies on a method similar to the one in the video (it is not hard to do so using the substitution to z-scores); that is, the method using the normal pdf still does everything in the video because the primary information--the fact that area under normal pdf is one--can be seen as the result of the multivariable calculus procedure. Nevertheless, it is an excellent shortcut to use, especially if you are pressed for time. Thank you for commenting the method!
excellent british pronunciation london or manchester city the accent ???
Or just using the generalized form of the gaussian integral
show off
The area of the normal distribution curve being 1 is a consequence of the solution to the Gaussian integral, not the other way around. In fact, the normal distribution IS a Gaussian curve, with a normalization constant. That's why it's called the Gaussian distribution.
when you were at integral e^(-2x²) you could use u=sqrt(2)x and it would turn into the literal gaussian integral of which you already know is sqrt(pi). Of course for the substitution you'd divide by sqrt(2) to account for the chain rule, so you'd end up with sqrt(pi)/sqrt(2) = sqrt(pi/2)
i had been messing with this integral for the whole damn day and all i had to do is complete the damn square -_- the rest follows up quite easily.
Thanks for this, im pretty hooked on mit integration bee integrals.
You can also do a u-sub : u = sqrt2 *(x+5/4) (it comes from the canonical form of -x²-5x-3). The bounds of the integral do not change and it gives the Gaussian integral again 🫦
I've watched some videos of Gaussian Integral and no one had explained as you did. Really good!
Okay, that's it. I'm going to start using stars instead of x'es from now on. It's just that funny to me, and don't ask why.
Excellent explanation on evaluating the Gaussian integral. I like the visual aids you used to highlight the intuitive discussion and the three comparisons made between Cartesian and Polar coordinate systems. You did a great job! Thank you.
Was able to do this on my own. Same method as it is in this video. All the binge watching of integration videos is paying off.
I m an IIT JEE aspirant.
So We have only learnt simple integrals only.
But you have explained it very good.
I was having curious of integrating e^-x^2.
great strategy! The fact that range is from -inf to inf makes this problem handleable
At 4:00: x and y are not just some random variables, they need to be Cartesian coordinates for the spherical coordinates to work. That puts limitations on the Gaussian integral because it would not work in singularities.
I love these videos, your explanations are very clear and thorough even for people who haven't seen this type of integral before but for those who have It doesn't take long to solve If you know the Gaussian Integral well, 1)Complete the Square 2) pull out the constant 3) Apply u substitution and you're done. Took me about a minute.
Omg thank you so much for this intuitive explanation of the Gaussian integral, my understanding has grown exponentially!
Thanks for the video, it really helped me understand polar coordinates (something I never actually understood). I wouldn't have had a clue on how to do this kind of problem before. Your solution was really nice, but after watching it i noticed that you could've made it simpler after you got to the integral of e^(-2x^2) dx by using u-substitution. If u^2=2x^2 then 2x*dx=u*du, dx=u/(2x)*du, dx=1/sqrt2*du. This makes 1/sqrt2 * int e^-u^2 du which gives sqrt(pi/2)
excelent i do love mathematics
i just finished calc 2 and on my way to 3. still it made some sense.
Excellent! This is the best explanation of this I have seen out of four. Very well done, sir! I have subscribed!
Excellent job oh my god, awesome
Thank you, joanmartin suarez loaiza! I appreciate it.
You are doing wonderful job, even though I am not a science or maths student but I like watching your videos ,please make videos on probability , permutations and combinations.
Thank you mister for this considerable amount of knowledge😇😇😇
Of course! Thank you, Samarth M.shetty, for an encouraging comment!
17:20 why don't you go with (x × sqrt(2) + 5sqrt(2)/4)^2 to make the exponential directly become -u^2 and not -2u^2 ??
you got the best explaination so far.God bless you
Omg best channel for integration
This is such an enjoyable explanation : )
Well done. I could not solve it on my own. Thanks for sharing.
Nice. It would have never occurred to me to approach the problem that way.
I think it should be clarified why the equality @5:45 is allowed. It normally isn't allowed for the same reason you can't multiply different summations together. However, in this case, because the outcome of the integral in question converges, ie the outcome is a constant, and because you can pull constants in and out of integrals, you can simplify it into one integrand as long as the variables being integrated over are different.
There's a pretty good explanation here: math.stackexchange.com/questions/549923/how-the-product-of-two-integrals-is-iterated-integral-int-cdot-int-iint
in last section you should have just change of variable v=sqrt(2)u
Sorry, I'm not really familiar with "MIT integration bee", but how long do you have time to solve it during the competition? I mean, solving this in maybe 3 minutes or so, would be absolutely insane.
I agree. The only way you can resolve this on time is already know the gaussian integral very well.
gaussian integral is REAAAAAAALY important at statistics.
You are given 20 minutes to solve 20 problems. IIRC
Kosekans
Well, if you consider that he "wasted" 15 minutes deriving the Gauss Integral (he even explained polar coords and how to evaluate the Jacobian) and some useless steps afterwards that he did to make things look easier and more evident, this integral can actually be evaluated in 2 minutes, maybe less:)
Needless to say, the Gauss Integral is very famous and extremely important in statistics since high school
This would be a complete the square and you’re done kind of problem. No actual calculus just recognizing patterns in the math and going off of it.
That's very helpful. Thank you so much!!
fantastinc solution easy understanding as a little boy
It can be solved much easier, but this solution uses Gamma function.
Hand clapping emoji! Realy nice, thanks
You made that easy. Thank you.
You have made it beautiful. ❤️❤️❤️❤️
Congrats on the explanation. It was very helpful.
I'm in Grade 11 and I really want to learn things like this but we are limited to the curriculum can you maybe consider creating "math for dummies" videos pleaaaaase I really do enjoy maths
Claudio Gomes I'm just in 7 grade and I also want to learn things like these. Though the curriculum is limited for a std. Who cares you do what you want. And your idea is great dude
@@kesavareddy9791 Khan academy.
Claudio Gomes Search 3blue1brown
Thank you teacher 🙏
You can do it with only substitutions knowing the antiderivative of an exponential and the answer to the Gaussian integral oh yeah and some algebraic manipulation
At the end, why didnt you say, u = sqrt(2)*(x+4/5)?
Then you could use what you found out before becuse you would have e^(-u^2)
And you would end up with the same solution
omg I was literally just looking at this 5 min ago
That was amazing, congrats
Excellent video!
Could you solve this integral if the limits were from 0 to infinity?
5:33 your ex is going from neagtive ienfinity to infinity lol.... i laughed badly in this
Hmm I wonder what is integral from -∞ to ∞ of e^(ax^2+bx+c)
The method in this video doesn't work for that. You would need the integrand to be of the form e^(-ax^2 - bx - c), for which the integral, using the method in the video, evaluates to sqrt(pi/a)*e^(1/4a)(b^2 - 4ac).
Abdullah Almosalami Thank you. Where’s the proof for this?
Radio TV Using the same method as in the video. So, given your integral is of the form int(e^(-ax^2-bx-c)), you do the following:
Factor out e^(-c) since it's constant
= e^(-c) * int(e^(-ax^2-bx))dx
Now rewrite quadratic by completing the square,
= e^(-c) * int(e^( -a( x + b/(2a) )^2 + (b^2/4a) ))dx
Now factor out e^(b^2/4a) since it's also a constant and simplify with e^(-c) from previous step,
= e^[ (1/4a)(b^2-4ac) ] * int(e^[-a( x + b/(2a) )^2)])dx
Now setting I = int(e^[-a(x + b/(2a))])dx = int(e^[-a(y + b/(2a))])dy, and keeping in mind both x and y go from -inf to +inf, and simplifying resulting expression,
I^2 = int(int(e^[-a( (x + b/(2a))^2 + (y + b/(2a))^2 )))]dxdy
Now, to make the substituion of r^2 = x^2 + y^2, so we can convert this definite integral to polar, let's the make the substition u = x + b/(2a) and v = y + b/(2a), which now makes the definite integral,
= int(int(e^[-a( u^2 + v^2 )])dxdy
And now we make the substition r^2 = u^2 + v^2, and dudv = rdrdt (I'm using t here for theta) as shown in the video, and keeping in mind the bounds of the integral are now 0 to 2pi for t and 0 to inf for r, and so,
= int(int((e^[-ar^2])r))drdt
And now using substition (kind of running out of letters lol) k = -ar^2, dk = -2ardr => (-1/2a)dk = dr, and since the bounds of r were from 0 to inf, the bounds of k (since we set k = -ar^2) will be from 0 to -inf, and so the integral is now,
= int(int((e^[k])(-1/2a)dk)dt)
Now rewrite the double integral so it's a little more clear,
= (-1/2a) * int(dt) * int((e^k)dk)
Solving, remembering the bounds for k!
= (-1/2a) * 2pi * (-1)
= pi/a
And now, remember that this was I^2, so
I^2 = pi/a => I = sqrt(pi/a)
This, btw, is the solution to the general form of the Gaussian integral int(e^[-a(x + b)^2])dx with b instead being b/(2a) which doesn't affect the result. Now that we have I = sqrt(pi/a), the original integral was e^[ (1/4a)(b^2-4ac) ] * int(e^[-a( x + b/(2a) )^2)])dx, and we just found the write part of the product, which is I, so the solution is,
e^[ (1/4a)(b^2-4ac) ] * sqrt(pi/a)
Abdullah Almosalami So cool, thank you!
try to form a pdf of normal distribution, then it's much easier
Hi, I'm thinking if I should take degree in Math and applied math or electrical engineering. May I ask what you are doing right now?
Loved your voice and teaching method
Thank you Sooooo much
16 and 18 is also pretty confusing...
Of course! I will try to look into 16 and 18 in my free time in the near future, but I cannot promise that there will be videos on them. Thank you for your recommendations!
Set speed to 1.5
Great
You've got like 20 accents mixed up
Smooth af
you forgot to rationalize the numerator at the end
Hey, pretty cool that Gaussian integral :DD
But what makes you feel so sure it's I = sqrt(pi) and not I = - sqrt(pi) 🤔
That made me a little confused, but everything else was just perfect, congratz for being really didatic, you'll go far that way xD
Since e^(-x^2) is always positive on the real line, the area under the curve should be positive. =)
@@LetsSolveMathProblems Oh, I see, thanks :DD
Oh come on, I thought you would’ve given us more of a challenge! I could solve this easy if I just u subbed x with -(4x+5)/2/√(2) (a simple translation and dialation), leaving behind a constant times e^(-u²), which can be easily integrated using the error function, which is solvable for +-∞. Realistically, the hardest part of this would be figuring out what to substitute u (which isn’t that hard anyway if you remember -b/(2a))
19:54
But x≠u
How can you do that?
Pls someone explain me
You're confusing two different things. At that step, he's not making a substitution based on a relationship between x and u. He's saying the same thing that he did at 3:29, that the variable is an arbitrary dummy variable and the value of the integral is the same regardless of what we call it. i.e. It could be integral of e^-2(star)^2 d(star) instead of u and du (or x and dx) without making any difference.
oh my god
your pronunciations are excellent !
You have a super cool accent!
Thank god I chose med school
would have been way faster gaussian=(π)^(1/2)erfc(0)=(π)^(1/2)
Could you do question 15 of 2017 MIT integration bee qualify?
I am planning on doing so tomorrow as long as my schedule stays open! I worked on the question as soon as I saw your comment, and I believe I have an elegant solution. =)
LetsSolveMathProblems Thanks！
i only dont understand why dxdy=drdO*r, why not dxdy=sqrt(2)drdO*r
As stated in the video integrating over dxdy is best thought of as "adding up tiny rectangles". The same concept applies when we integrate in polar where you can think of adding tiny arc sections which might look like curved rectangles where the "width" of the rectangle can be easily thought of as dr, the base would not be simply dtheta obviously as that would mean an angle representing a length but instead is thought of as rdtheta where theta is in radians due to the definiton of arclengths (where pi came from ex: a circle of diameter 1 has a circumference of pi). If you were to instead use root2*r*dr*dtheta you would need to divide dtheta by root2 in order to get the right arclength (i know dividing dtheta doesnt make much sense, using a sensible angle such as pi/2 is easier to think of). Essentially it is because we use radians, as i said before using another angle system being the angle in radians divided by root 2 would make this work out
So his definite integral would end up being from 0 to 2pi/root2. Super long reply but i hope it kinda clears things up, if not maybe find a good review video on arc lengths and the unit circle. Unit circle helps with all kinds of intuitions so its worth reviewing anyways.
I am from Ukraine High School and i dont undenderstand english , im sorry for your 2 big comments
@@kavynizde no problem, big idea was unit circle :)
This guy has phd in integration....
It's star not s-tar
Se volvió un culo al final.
Dereveteb
Why m i watching this , i am just 11 years old
14 /15 now 🙃 Btw Calculus is fun
youtube education
너 한국인이지
나도 이 생각했지
u need to ask viewers to subscribe to the chanel. as per what i can see your videos have like 3/4k views but u only have 1k subs. that does not make sense. u are very good as math. now get better at selling.
Your comment is one of my favorite comments of all time! You are absolutely right--My abilities as a salesperson have never been outstanding. I will consider doing more advertisements in the future. Thank you for your encouragement and suggestion! I appreciate it. =)
Yeah he is right u r awesome
rdr golfish.
I don't understand any of these and Idk why it's in my recommendations
INT-A-GR-UHL! NOT SAYING IN-TE-GRAL
Funny accent
Why not just complete the square? It's way easier that way.
It's IN-tegral, not in-TEG-ral. Ugh, ruins this video.
Wow that's unnecessary, I think it's pretty understandable albeit the awkward pronunciation, and in no way does it hinder the video's main point in any manner. I sincerely hope the teachers you've had in your life didn't find you as much of a dotard as you seem to be
The way you pronounce integral is triggering me
thank you very much