How Gauss solved the integral of e^(-x^2) from 0 to infinity 

Same

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

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

that proof is cool but beyond the scope of undergraduate

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

​​@@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.

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

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

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

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