Integral of exp(-x^2) | MIT 18.02SC Multivariable Calculus, Fall 2010

Integral of exp(-x^2)
Instructor: Christine Breiner
View the complete course: ocw.mit.edu/18-02SCF10
License: Creative Commons BY-NC-SA
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[@Deshammanideep]

mathematics is more beautiful when you are not studying for exam but learning for fun.

[@MohOEM]

Yes, I completely agree with you my friend. My grades in mathematics courses at university were bad. But I always had a different look at it. It's always fun!! Specially, when geometric interpretations are cleared out.

[@gabrielmachado146]

Your quote should become a proverb... That's exactly what I think too!

[@sandeep2435]

Studying must be for knowledge and for breakthrough but not for exam

[@breathtakingsamurai981]

Exactly! The education system is so fucked up that it makes students to hate science because of those useless exams, grades and other shit.

[@hattorikanzo2793]

When you are learning for fun, you should never have a problem in getting grades. I guess its a common thing these days to blame the education system

[@Supadupatight]

Just finished Calc I and watched this video for fun but just wanted to say I loved the clarity of her explanation!

[@GuillermoRobles]

Jennifer Pena if I =1 then Isquared is 1

[@CandidDate]

does this relate in any way to the error function?

[@deekshithnaidu4871]

I wonder why this comment got 98 likes ? XD

[@tibetannoodles]

just calculus 1 wont help you understand it

[@mr.knight8967]

Math QUESTION Integration problem kzbin.info/www/bejne/q6WUnmSco75kgrs Best problem watch it .

[@BrunoFerreiraTheOneAndOnly]

The last time I solved a double integer was in 2018. This woman literally saved my life right now. That's exactly the integer I wanted to solve but couldn't remember how to. Thank you!

[@StNick119]

*integral

[@raunaqchopra5951]

Her ability to explain never ceases to amaze me. Just legendary!

[@pipertripp]

Who the hell figured this out orginally?!? Genius.

[@camposcuanticos]

With this method, Laplace.

[@leguaani1234]

Was thinking the same, absolutely genius answer

[@stephenbeck7222]

If the paper from York University (UK) cited by wikipedia's article on Gaussian Integrals is accurate, then Laplace actually used a method with rectangular coordinates that came down to an arctan anti-derivative. His contemporary Poisson is cited for the slightly simpler method shown here with polar coordinates.

[@sequorroxx]

This is what separates math users like me from mathematicians. I can use the rules from one step to the next, but I have no imagination, no vision, no creative insight into which path to take to get from A to B. Mathematicians see the way. I can only follow.

[@meleneleneme4545]

there are actually 11 more ways to solve this..

[@victorvega8061]

I swear to jeebus this is easier to understand than transformations. Amazing how it just takes a small change in perspective to make this problem easier!

[@bendavis2234]

This example really blows my mind. It surprised me that adding another variable made the problem solvable in the end. Also, the explanation was very clear!

[@Luiferhoyos]

Thank you so much, and the subtitles were so useful, because my listening is not so good at english.

[@mr.knight8967]

Math QUESTION Integration problem kzbin.info/www/bejne/q6WUnmSco75kgrs Best problem watch it .

[@ΔημητρηςΓαλάζιος-ο2δ]

This one is very important for physics (statistics). 90% you have to solve it in a project. Nice method.

[@IDoNotFeelCreative]

She's my favourite person in the world at the moment:D

[@sublimetrance]

I love these MIT videos... really helps remember some of these tougher concepts.

[@MrArtbyart]

I didn't expect a simple easy evaluation within just three minutes

[@xavierplatiau4635]

Did this exercice back when I was 18, there was 1 hint which was « First show that I = sqrt (Integral from -oo to +oo of exp(-x^2-y^2)dxdy) » Then I must say the change of variables x=r cos(t) y=r sin(t) is rather obvious.

[@bsul03420]

I really enjoy Christine Breiner's lectures - they are always the epitome of clarity.

[@sugongshow]

Dr. Breiner, thank you so much. If you ever come to Japan, please let me know, and I'll be glad to show you around.

[@WantedDeaDorAIive]

this is proper teaching! what an amazing video Clear concise explanations, no steps missing in the logic, every step is justified. This woman is one of the best math teachers i have ever seen.

[@hzkzg1614]

Hands down quickest easiest explanation out of the 3 videos I've seen on this today

[@gausiano3121]

Very good. I know three ways of solving this integral, by the method you did, by viewing it like the area under the gaussian PDF and by using the gamma function.

[@deepakpizar5830]

6:58 was awesome. I never thought like that

[@sheffielddu4803]

This’s awesome. Change a single variable problem into a multi variable problem, then find the equation ,and then change into polar coordinate ¡

[@094_f_shatanshubodkhe6]

If you substitute x with ax then multiply both sides by e^(-a squared) and then integrate zero to infinity , can directly give the desired answer

[@owen7185]

Some people are born to be good at lecturing and she's one of them

[@AJ_42]

I'm glad I'm revisiting my math. It's fun and almost recreational now, even though in college it felt like running a race.

[@pasodirect]

Tisztelt tanárnő, eddig ez a legérthetőbb Gauss integrál levezetés, gratulálok. Sokat segít a feliratozás.

[@sergeiohezin1219]

Самый краткий и простой метод вычисления. Супер!

[@dylanparker130]

Ah, this was beautiful. Can you imagine the joy of whoever first thought to do this?

[@akhshitabhat3883]

Thank u ma'am, came here after watching a lot of videos, but this one was really awesone enough to make a viewer complement this to u💙💙

[@anupam1095]

*If we use gamma function* ∫e^-(x²) dx, limit -∞ to ∞ let F(x)=e^-(x²) then F(-x)=e^-(x²) id est(i.e.) F(x) is an even function. 2∫e^-(x²) dx, limit 0 to ∞ put x²=z. 2/2∫e^-(z) dz/√z, limit 0 to ∞ ∫z^(1/2-1)·e^-(z) dz, limit 0 to ∞ equation converts into GAMMA function. Γ(1/2) = √π

[@marcodavicino414]

Hi everybody, for those who don't know what she did when she integrated e^(-r^2)r. She chosed -r^2=u, so du=-2rdr. It's a substitution.

[@exlife9446]

i know the guass blur filter in photoshop use this distribution, but for accelerate reason, we donot compute on a square or circular area of pixels nearby the center point, that will be very slow. instead, we do 1-dimension guass blur two times, because the curve is symmetry and same in all directions, so we can do it in two independent directions (when we zoom a image to a different resolution we can also use this method ). first for the horizonal direction, got a middle result, and then do the vertical direction blur on the middle result, we got the final result. also, on mathmatics, the curve covers infinite pixels, but in computer science, we just use a limited range of pixels, that is pixels near the center in 3 sigma far away from the center. and the sigma in the formular in photoshop is called guass blur radius.

[@박주은-f4x]

정말 감사합니다 ㅠㅠㅠㅠㅠㅠㅠㅠㅠㅠㅠ 당신은 한 공대생의 생명을 살렸습니다!

[@dedicated2304]

Have you also got sucked in this question. Actually me too😅😅!

[@ll4ves457]

Thank you! I had to integrate that for my quantum class and I was lost. I much prefer this explanation as opposed to introducing the error function

[@jamesbentonticer4706]

Very well put.

[@xoppa09]

Very clear teacher

[@jeffreyanderson5333]

The best illustration of this problem on the web!

[@nichel7]

I was supposed to solve a very similar integral (e^(x^2 + y^2))dxdy, but I couldn't really find a way. It saved me a lot because tomorrow I have an exam and they usually take exercises like that, thank you very much for the great explanation :)

[@TheRumpusView]

An excellent lecture; clear and concise and delivered by a very sweet lecturer. Excellent combination.

[@SteveMcRae]

I remember watching this video before, and didn't quite understand it...but after watching George's video on it I think I will try to watch it again.

[@mmogib]

@therealgrillninja the function e^{-x^2} is positive for all values of x; therefore the integral over (-inf,+inf) is interpreted as the area under the curve y=e^(-x^2) and the x-axis; hence the rejection of -sqrt{pi}.

[@samruby82]

invaluable. this is what relates pi to normal distributions in statistics. what I still don't get is why most sources online integrate this to be: (1/2) * sqrt(pi) * erf(x)

[@markproulx1472]

Very entertaining. I’ve always struggled with mustering the insight to crack these types of problems.

[@theshadypersonify]

How is this even possible. I love this teacher by the way :)

[@Etothe2iPi]

That's exactly what happened at university over and over again. Spending a vast amount of time on absolutely trivial algebra and glancing over the important part - dx dy = r dtheta dr.

[@lukapopovic5802]

Etothe2iPi Well, that's whole another lection dx*dy = dA = r*dr*d(theta)

[@lukapopovic5802]

Etothe2iPi delta(A) in polar coordinates is aproximately a rectangle (as dr and d(theta) tend to 0) with sides r*d(theta) and dr. You need to see someone drawing this, it's hard to understand without looking at a picture of this.

[@Etothe2iPi]

Luka Popovic Good idea. A picture sometimes says more than a thousand calculations.

[@ib9rt]

I paused the video at that point, grabbed a pen and paper, and worked it out for myself. That is what learning is all about. If you wait for someone to give you all the answers you will not grow, you will not develop your brain, you will not advance. You will leave university the same as you entered.

[@MultiChantal22]

She was not glancing over the importance of dx.dy=r.d(thetha).dr, it was already taught before

[@arctan-k]

One of the most beautiful integrals😍

[@brianjohns49]

Thank you! Great lecture on integration. I think its integrated to your viewer's minds very well. :-)

[@jakolu]

who is this lady? She is great at explaining things. I am quite drunk yet I can still understand what her quite well!

[@snehalsha6476]

"Bartender , 2 shots of vodka and side of integrals"

[@pbj4184]

4:44 Mind blown! Had never thought you could do that

[@JimHaroldson]

{The way she did it in the video is the only way I have seen anyone successfully solve this particular integral.}

[@notbnull]

this is insane

[@asparkdeity8717]

Meanwhile in my Uni maths course, this integral was briefly stated as a 10 second long remark after learning about Jacobean transformations and n-dimensional integrals in our vector calc course, explicit steps like this were never shown

[@Atrimodes]

Although it is true that the square root yields two answers, the positive one is the only one that makes sense. The function exp(-x^2) is a positive valued function only, so -sqrt(pi) doesn't represent anything. It has nothing to do with the bounds. The definite integral of this function, evaluated from 0->infinity is actually sqrt(pi)/2. The bounds have not changed. Sources: Wolfram

[@bzhell4544]

thanks for the clear explanation to get the right answer. appreciate it

[@mohammadaunaisbhat1944]

6:46 The substitution way is more correct as the way you have integrated e^(-x²) here at 6:46 we should have done this earlier, in the first step.

[@dimensionentangled4514]

when we change the variables, the limit runs from 0 to 2pi in one and in the other it should run from 0 to a (radius of the circle, since we are now in the (r, theta) coordinate)

[@xavierplatiau4635]

x and y going to infinity means that you must consider circles which radius go to infinity too during your change of variables.

[@davepastern]

Excellent tutorial Christine - clear as glass and easy enough for a maths idiot like me to even understand! Was very enjoyable to watch.

[@DanielCharry1025]

For the explanation of I^2 you must look for Fubini's theorem... For the r dtheta dr go to the Change of Variables Theorem. psst Jacobian...

[@positivegradient]

She is an awesome teacher.

[@m35926]

8:16 that's the most perfect pi created by a human ever.

[@themaverick1891]

You are wrong. It's not.

[@boricuababyhuey7576]

How do some people actually think Chemistry is easier than this?! This make logical sense and the logic is consistent without exceptions. Great professor!

[@edweinb]

Beautiful and easy to follow. Wish I had watched this video first!

[@اميرةابي-ص2ك]

شكراً ✋🇮🇶

[@oscarcanteromanrique1668]

Muchas gracias por su video, espectacular y sencillo.

[@rajendramisir3530]

Gaussian Integral. Normal distribution function. I wonder how Carl Gauss found the anti-derivative of this integral. Christine, you used a clever method to evaluate this integral.

[@WoobyMe]

I felt like I was the one this was made for, I needed this for my probability theory project. Thank you so much!

[@akashmajeed3911]

Its Awesome to learning from you teacher

[@pandit-jee-bihar]

This whole jacobian thing and double integral was the easiest thing in the engineering maths provided you remember the trick.

[@SachinVerma-br3ho]

what's the trick?

[@PreciousMental]

@@SachinVerma-br3ho lol get bodied (I don't know "the trick" either)

[@fabslyrics]

things of beauty :) i was amazed the first time i saw this on the benches of the Sorbonne university in paris in 1996 :)

[@bjorksven]

Yes, it is a constant, approximately equal to 20. What value the integral has depends on your boundaries. You don't need to do what's explained in this vid to calculate your integral, because that's just a regular polynomial :)

[@alexandrklimov2601]

Люблю смотреть ваши видео жду новых выпусков

[@DaSoldat33]

This was very useful for solving Guassian normalization constant... Instead of I^2 I use I^-2 C is equal to inverse of this function

[@badgermcbadger1968]

Can you explain how?

[@were65were]

@donkaru1 You are integrating from -inf to inf, meaning all real numbers for both x and y. That means that for every x number, there is an equal y number. An easier way to see it is just by saying x and y are dummy variables. You could say the integral from -inf to inf of e^(-(a^2)) da and it would be the same result.

[@owen7185]

Best explanation I've ever seen, facts!!

[@pedrotorresperez4213]

Congratulations from Spain

[@AndrewSmall963]

This is counter-intuitive, as plotting this function against a unit circle [Google search y=e^(-x^2),y=sqrt(1-x^2) to display] makes the area under the function look far more than the difference between pi/2 and sqrt(pi) (approx 0.2 or 13% of pi/2).

[@shadon_official2510]

great instructor.

[@TrevorKafka]

@joddle23 You actually don't need the Jacobian to figure it out for the polar case. Let me see if I can describe this. Think about how to define an area element in polar coordinates. A small (slightly curved) rectangle will have sides oriented radially and tangentially. The tangential sides will be r∆θ whereas the radial sides will be ∆r. The area ∆A is approximated by r ∆r ∆θ, or in the limiting case, r dr dθ. It's because as you go further from the origin, the area element gets bigger.

[@nuocmat]

nice . I was trying to figure how to integrate this erf function in my wireless class but couldn't figure it out, so I use my calculator instead. but now I know how to do it by hand. thanks

[@exlife9446]

it is a beautiful function...

[@alimubarack]

Brilliantly explained

[@pranshuprajapati6090]

Your explanation is good

[@TheLivingHeiromartyr]

Well I didn't know that. There's no way of finding a nice closed form for the definite integral of e^x^2 between two finite bounds. You can approximate it by expanding the taylor series for e^x^2, and then integrating each term. But there's no elementary function expressing the integral you want. The same with this integral. We can only compute it definitely, and by estimation. the -inf to +inf is a convenient special case.

[@usman5954]

Watched. I liked it but a little hazy on how to deal with the Jacobian thing? I think you made reference to one of your previous lecrures.

[@jeremygraham7976]

She is awesome!

[@parvelius]

It's possible to evaluate the integral by using complex analysis as well. But I prefer this method.

[@charlesbromberick4247]

This gal has got a great Jacobian and makes duck soup out of this integral. Very clear and well done.

[@Sidd-rb4ec]

Made my day technically night

[@JimHaroldson]

I haven't taken that yet - but I imagine that is elegant. :D

[@AnnaHeinemann1986]

Thank you so much! You saved me a lot of time/work ;D

[@moinmalik1320]

Going to feel good for the rest of the day.

[@wadadira1677]

Excellent job and beautiful

[@jobsism10]

There is a slight error: SQRT (I^2) = +I only if the limits of integration are from 0 to infinity. So, essentially, the integral that you finally evaluate is actually the integral of (e^(-x^2)), FROM 0 to infinity [and not the indefinite one from (-infinity) to (+infinity)].

[@booli8542]

No.

[@yashsaraswat8932]

Lovely explanation 👌👌

[@adamanimations7525]

gaussian integrals are so helpful for these integrands but lebesgue integral is amazing for any type of graph

[@김현아학생이과대학정]

thank you so much my professor

[@two697]

Does anyone have know the video for what she was talking about in 5:55 where she changed dxdy to rdrdtheta??

[@vm667]

I totally got everything she said..

[@were65were]

@lukegranger89 she said it was going to be useful in statistics and if u had taken that class you probably know about the normal distribution which as you know is a e^-(x^2) with modified magnitude so that the integral is 1.... so yeah, get YOUR facts straight first

[@joddle23]

@Yakeyglee yeah, in our multivariable calc class a few days ago, the professor said that dA = r dr dθ (though he didn't derive it), and then I realized that a small portion of arc length of a circle = r dθ and therefore, a small portion of area = r dr dθ! I guess that's what you were trying to say, but yeah, now I can appreciate this video fully :)

[@eppid818]

This is a pretty cool video, and I learned alot in the process

[@nguyenhuuminh9588]

it's very nice solution From HUST-Viet Nam

[@teekayuk]

that is one amazing professor.

[@aadilansari5997]

Playing fast and loose with Integral operator is not recommended, one may not object to getting fooled just so that integration can be calculated "easily". This is how it is given in Thomas and Finney too.

[@virtualcat5581]

This is how I learned it.