To see how Laplace solved the Gaussian integral, click here: kzread.info/dash/bejne/pneEs7WoktSsmJc.html

Very cool integral. Seeing this helps me to understand why the function for the normal distribution curve has the 1/sqrt(2*pi) in the front.

I am obsessed with this integral and the gamma function

@dumbass4385

Ай бұрын

Same

Can't believe a semester ago I mastered this and now I completely forgot how to solve this. Thanks for showing me I have to practice more.

I find it interesting that the jacobian matrix you end up getting is just the rotation matrix in R2 multiplied on the right by the column vector [1 ; r]. There must be an explicit reason for this beyond "the math just works out that way."

He's truly the best

beautiful

brilliant

We get this problem in Class 12th Cengage Dpp

too good

The region of this double integral is a circle ,the bound for theta should be integrating from 0 to 2 pi , isn’t?

@bprpcalculusbasics

Жыл бұрын

Oh bc we are only integrating it from x=0 to inf. We are in the first quadrant.

Can you explain what the jacobian is and prove the formula to calculate it?

@user-ks5ci6bs6x

Жыл бұрын

that proof is cool but beyond the scope of undergraduate

@ShanBojack

Жыл бұрын

​@@user-ks5ci6bs6x true but an explanation would be nice

@oke5403

5 ай бұрын

​​@@user-ks5ci6bs6xreally? in poland i had it first in analysis 3(we don't have calc curses specifically but analysis is mostly the same as calc in america pretty sure) and then the use like demonstrated here in analysis 4. both times with proofs.

@lucasfranco1758

2 ай бұрын

if I remember correctly, there's a great video on Khan academy about this subject

I neeb to know the reson why the two therm integral combine togethe?

@Abedchess

6 ай бұрын

Integral of e^(-x^2) dx from 0 to infinity is just a number. Remember, it is a definite integral. Integral of e^(-y^2) dy from 0 to infinity is also another number, which is the same number as before. If you let I = Integral of e^(-x^2) dx, (Note, I is now a number also. I'm lazy to write out the limits, but just assume the limits are all 0 to infinity) and since you know Integral of e^(-x^2) dx = Integral of e^(-y^2) dy Thats why I^2 = (Integral of e^(-x^2) dx) (Integral of e^(-y^2) dy) = double integral e^-(x^2+y^2) dx dy The reason for this procedure is because, there is a neat trick for integrating anything containing "x^2+y^2". The trick is changing to polar coordinates. As for why integral f(x) dx integral g(y) dy = double integral f(x) g(y) dx dy. Its because of the property of summation

Sir, I don't understand the point of intigral multiplication? Plese shwo me

how did u know that it is in the first quadrant?

@advaykumar9726

7 ай бұрын

Both x and y lies from 0 to infinity therefore both are positive Hence it lies in first quadrant

Why integrate theda from 0 to pi/2 but not 2pi?

@__fahim.__123

8 ай бұрын

Ig its bcoz we jus only considered the first quadrant in the circle when we changed it into polar form like more than pi/2 its juss a matter of perspective

@mitcigamer4289

3 ай бұрын

the original bounds of both x and y were positive and hence we're only in the first quadrant. Hence the theta can only be 0 to pi/2

-infinity to infinity π

First

That is not correct manner , because the y= e^-x^2 is not equal the e^-y^2