Believe in double integral, NOT single integral 

dude Your photo... =(

I like the one with 2 markets =(

Awoms video Also

Thank you so much for you nice comment! I am very happy to hear this!

Will you please make a video to find the coordinates of the points of intersection of two intersecting circles.

@@aniruddhaghosh1303 thar is a nice tho Mmmm, well, a clue that I can give you is that, don't think on a function world accepted by school Just, plug the same varibles and use the quadratic formula for find out y [f(x)] That is something that works out

Distance formula at certain coordinates?

@@junkgum mmm Sqrt[(y2-y1)^2+(x2-x1)^2] If I am not bad

Thanks!

Love you bro

Glad to help!! 😃

What exactly would I have to prove here to be able to use it

@@alphazero339you’d have to prove(or already know) that the function f(x,y) is integrable in the product measure space. Meaning that when you integrate |f(x,y)| over XxY with respect to the product measure on XxY(these things have to be properly defined using measure theory), then this gives a finite value. Then we can take the integral of f(x,y) (without absolute values) over the product measure space and evaluate it as a double integral and exchange the order in which we integrate. In practice, we can more easily use Tonelli’s theorem here: If f(x,y) is non-negative and measurable then we always have this equality, but the integrals may not be finite. In the video’s case, the exponential function is always non-negative and is measurable so this works and no need to even verify Fubini!

Are you differentiating the numerator and integrating the denominator?

@@tommyliu7020 Yes, I did a first IBP letting dv = 1/t^2 and u = [1-e^(-t)]^2; then I've repeated this procedure with the resulting integral, obtaining -2 ∫_0^∞ [2e^(-2t)-e^(-t)] ln⁡ t dt

Why are you telling your grade

Yes, as the function is nonnegative and measurable, and the exponential function is nonnegative

Integration by parts gives me 2\int_{0}^{\infty}\frac{(1-exp(-t))exp(-t)}{t}dt Integrand and interval of integration hints me to use Laplace transform so i calculate Laplace transform L((1-exp(-t))/t) plug in s = 1 and double the result To calculate L((1-exp(-t))/t) it is enough to calculate L(1-exp(-t)) and integrate the result

Only in the cases when it's not pi²/6 or the Euler Mascheroni constant. ;)

@@bjornfeuerbacher5514 Indeed.

Yes bc those integrals go from 0 to 1

@@blackpenredpen ahaaaa tyyy

Could you please provide me the time stamps? Thanks.

@@blackpenredpen It's pretty common occurance. Just in this video: 0:00 - 0:05 about 7 times 0:14 single one 0:33 single one 0:48 - 0:50 a couple And so on. It sometimes seems to correspond to movements of the mic, like the slight upward hand movement at 0:05. Could be a faulty connection or frayed wire.

Thank you so much for pointing those out! I will see what I can do to fix it!