This isn't a Circle - Why is Pi here?

this shape is known as a bell curve or we might call it a gauss curve it's super important for many reasons one of the most popular applications is the normal probability distribution this symmetrical shape describes a lot of phenomenon in the real world and the way we use it to find probabilities is by measuring the area under a curve between different values we have to account for some normalizing constants but really this shape comes from this function e to the minus x squared and you know if we want the area under this curve well we're going to have to do some integration and this function i don't know how well i've drawn it but it stretches forever to the left and to the right it has an infinite domain so we want to integrate this thing over its entire domain to find the area so this is now our problem the definite integral from minus infinity to infinity of this function e to the minus x squared dx and what you may notice is that none of the regular integration techniques work very well here there's no like u substitution or integration by parts that's going to be able to solve this and we could use a power series but that's going to get a little bit clunky that's maybe how we would do it if it wasn't a definite integral if we didn't have any bounds so here's the trick here's the technique we're going to just call this i i for integral and i'm going to square both sides of this equation i'm going to square this entire integral not something you're used to doing but what does something squared really mean well it means times itself so this just means the definite integral from minus infinity to infinity of this times itself now a subtle trick here that this quantity it doesn't really matter that i called it x right we could have used ts or z's or y's and i'm going to use a y here to represent the other one and it's equi it's completely equivalent because when we actually integrate this well all the x's go away don't they think fundamental theorem of calculus we end up just plugging in for x so it doesn't matter what letter we use here i'll just call y on this second integral and these are being these are being multiplied but it's the same quantity they're both i so this is i squared and actually since both the letters are completely different here and they have the same definite bounds here minus infinity to infinity this is what's known as separable these integrals are separable if you go into multi-variable calculus you might try to split up these integrals to make them easier here we're going to combine them so this is exactly the same thing as a single definite integral putting these both inside the integrand so they're both base e when we multiply like bases we add the exponents so minus x squared minus y squared that's the same as minus x squared plus y squared in the exponent and then we have i really should need two integrals here we're integrating twice we're putting these together and i'll just say d y d x if you've had some experience in multi-variable calculus you might know where we're going next this is still a pretty wonky integral to solve but it's in rectangular coordinates that's making it wonky a square coordinate system but x squared plus y squared think pythagorean theorem think trig think the fundamental trigonometric identity x squared plus y squared is r squared the pythagorean theorem we're going to make a change of coordinates here we're going to switch to polar coordinates instead of using rectangular we will use the trig functions and the pythagorean theorem to change this coordinate system and i guess you could call it a substitution i mean really what we're doing is saying x squared plus y squared equals r squared we're going to have to change these bounds and in the polar coordinate system d y d x or d x d y this turns into what's known as its jacobian r d r d theta so we're changing from x and y's to r's and thetas and what we also need to do is change these bounds so if you think about the polar coordinate system we're instead thinking of not rectangles but circles the radius r starts at the origin and extends forever well originally the x boundaries when we think dx were minus infinity to infinity river to the left and forever to the right well for r boundaries the r always starts at the origin namely at zero and then in this case continues out forever so these bounds the r bounds will be zero to infinity and the y bounds which were minus infinity to infinity for the same reason turn into theta boundaries well if you think about the unit circle or just any circle we get coterminal angles so we can describe every single angle in one sweep of a circle 0 to 360 degrees or we typically use 0 to 2 pi radians this is the substitution and so at this point you might think oh we're good to go well we actually have to make one more substitution if i just gave you this integral in a calc 1 or 2 class you could probably figure it out via a u substitution letting u equal the minus r squared that would make d u minus two r d r and that's going to take care of this exponent and this extra factor of r because our dr is going to be a d u over minus 2 r so if you go ahead and make this substitution this outer integral won't change still 0 to 2 pi but this integral which was our boundaries we need to change them to u boundaries i like to make a little t table here for u's and r's originally r was infinity and zero well if we plug these into our substitution if r is zero u will be zero so this lower boundary stays the same and if we plug in infinity well you can't really plug in infinity but if we could think order of operations infinity squared would be infinity and then negative this would be negative infinity and we should line these up just as they are and maybe this feels a little bit uncomfortable typically we like the lower limit of integration to be less than the upper limit of integration we'll fix that with this negative that comes out of the substitution here because our dr that's becoming a d u over minus 2 r we still have a d theta we still have that r and e to the minus r squared became e to the u well thankfully now we're getting these r's canceling out and we can even take this negative and one property of definite integrals if you multiply by a negative it has the effect of changing the bounds or flipping them so really we could rewrite this definite integral 0 to 2 pi minus infinity to 0 e to the u d u d theta over two and we're finally ready to actually compute this thing if you don't mind i factor out the one half and we could even unseparate this now kind of doing the reverse of what we did before there are no thetas inside this integrand so maybe you would want to just make this 0 to 2 pi of 1 d theta integral minus infinity to 0 e to the u d u and compute these separately looks like we've got one half the antiderivative of one d theta is simply theta and we're evaluating from zero to two pi and then this is being multiplied by the anti-derivative of e to the u is e to the u evaluated from minus infinity to zero now i'm being a little bit generous with my notation here we can't really evaluate at negative infinity but what we can do is take the limit so traditionally what we would do is let t approach negative infinity here i don't have enough space on the board but just know that that's really what we're doing we're really taking the limit here when i plug in minus infinity so it looks like we're getting one half if i plug in 2 pi that'll be 2 pi minus plug in 0 that'll be 0 and this is being multiplied by e to the 0 which is 1 minus and again i'm not really plugging in infinity but if i could it would be e to the minus infinity which tends towards zero think of the graph of e to the x has that horizontal asymptote along the x-axis as we go all the way to the left e disappears so really this is like one half times two pi times one or simply pi wow did you expect a pi to fall out of here and we're not quite done yet because if this equals pi this was the square of what we actually wanted so really if we take the square root of both sides well we'll get what we want we'll get our original integral quite amazing that the area under that curve is square root of pi and this is where if you've ever seen these normal curves and the formulas for them you'll see a square root of pi or square root of 2 pi when we kind of adjust for some things a pretty amazing result from a pretty amazing integral now if you want to see another amazing thing click the video on the screen right here it has some info in there i think you'll enjoy i'll see you in that one

Channel: BriTheMathGuy

Views: 229,438

Keywords: math, maths, integral, integrate, guass curve, normal curve, normal distribution, probability distribution, gaussian curve, gaussian distribution, bell curve, bell shape curve

Id: S79KPrIm_Gc

Length: 10min 30sec (630 seconds)

Published: Thu Mar 04 2021