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

[donkaru1]

thank you,,but i had to ask,,u might seem that this question of mine is utterly trivial,,but i want to know ,,,if u treat (e^-X2) as I...and (e^-x2-y2) as (I^2),,,are u implying that (e^-x2) equalls (e^-y2)?...in that case u are saying x=y,,,but how can x=y for all situations?,,,that part does not seem logical to me,,,im sorry but pls clarify

[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.

[kulza23]

sidewalk chalk!

[therealgrillninja]

Can {I=-sqrt(pi)} as well? Since I^2=pi is there any reason "I" must be positive?

[chichoos07]

@therealgrillninja its an area ..must be positive

[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}.

[Dmaar]

nice and thank a lot

[dfjndsk2008]

took calculus so many years ago, forgot how to turn X Y into the polar coordinates

[lukegranger89]

exp(-x^2) clearly isn't a distribution because it doesn't integrate to unity! get your facts straight!

[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

[TheFlowerofFaith]

I love you soooo much! Great explanation!

[PmQable1]

multivatiable calc professor doesnt know how to do a u stubstitution...

[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)

[originalrhombus]

@samruby82 Because 2/sqrt(pi) int 0 to x ((e^(-x^2)) is the Gaussian error function, and if you multiply it with 1/2*sqrt(pi), you get [drumroll] int 0 to x ((e^(-x^2)). In other words, you get exactly what you started with. The moral of the story: don't bother trying to solve this with WolframAlpha. You have to do the symbolic legwork yourself.

[samruby82]

@originalrhombus hey great, thanks for the answer! Any idea why these computer algebra systems add the erf component needlessly in the first place?

[joddle23]

This is awesome! I only know basic multivariable calculus though, so I don't understand the Jacobian if that's what it's called... (how dxdy becomes rdrdθ with the change of variables). Oh well :P 

[Yakeyglee]

@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.

[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 :)

[Suburbanwolf]

Big help. Thanks!

[192ali1]

Thank you

[davidwb7777]

Excellent video. Thank you.

[borismilner]

This is really awesome.

Thank you MIT =)

[mathrock87]

integrate e^(-x^2-x^-2) how do ı fix:(

[tupacsingh90]

that is cool



[mtcjoseph]

Mary me!

[mtcjoseph]

@mtcjoseph Guess I don't stand a chance if I can't spell.

[LeoRockShady]

MIT vids have only 2 comments .. But Lady gaga's ones have more than million

[affablelochan]

thank you for the video. It was helpful for me while solving the expected value of a function.

[arlil1]

Excellent Video!

Clear, concise... great ;)