In this video, I showed how to rewrite and compute an improper integral of both types.
Пікірлер: 25
@user-xw6ky8ob4l4 ай бұрын
This Guy is Mathematician par excellence for all learners.Master of chalk and talk, respecting traditinal values of teacher and the taught rare example of fleeting era.
@chengkaigoh51014 ай бұрын
Incredible that a line of infinite length encloses a finite region
@cadenpink3164 ай бұрын
Wrap an infinitely thin string around your finger. You can wrap it around as much as you want, but you will never full cover your finger.
@gp-ht7ug4 ай бұрын
Isn’t there a little mistake when you put back sqrt(6)? Check the signs. But at the end the result doesn’t change
@Tomorrow324 ай бұрын
SQRT( number) is always positive.
@AquaticWaters4 ай бұрын
No yeah you’re right- it was supposed to be a negative when he brought the numbers down, but in the end it didn’t matter since +/- 0 is still 0
@joaomane4831Ай бұрын
That is not what they meant... @@Tomorrow32
@glorrin4 ай бұрын
Hello there, great video as always. just a small mistake that didnt impact the answer. On the one before last blackboard I* = sqrt(6) [ lim t->0+ [missing - here] tan-1 sqrt(t/6) + lim t-> inf tan-1 sqrt(t/6)] sinc lim t->0 tan-1 0 is 0 it doesnt matter if it is + or - but still. Also missing a * on the very last line but that is insignificant.
@antonionavarro10004 ай бұрын
Hubiera apostado la vida a que la integral no convergía, por el parecido de su gráfica con la gráfica de 1/x. Pero no, me equivoqué y efectivamente converge a √6/2•π Gracias por el ejercicio.
@wolfwittevrongel80674 ай бұрын
The tumbnail is wrong tho, great video
@PrimeNewtons4 ай бұрын
Thank you. I fixed it
@Annihilator-014 ай бұрын
Thank you so much ❤
@iquesillos124 ай бұрын
Amazing!!
@tomctutor4 ай бұрын
I notice that the integrand 3/[(√x)(x+6)] has no roots in the ℝ domain. So the integral is indeed the area under the curve. You did not mention areas so not an issue here; but if it were the area you were calculating then you would need to check for roots first. (A lot of students forget to do this and just blindly assume the definite integral of a function is equal its area). 😁
@samueldarko936013 күн бұрын
Hello, I dont understand why you substituted with root 6 tan rather than just tan theta, why root 6
@saarike3 ай бұрын
Simply Great!!!!
@alifiras11304 ай бұрын
Can i solve the integral by using partial fractions?
@PrimeNewtons4 ай бұрын
Try it.
@nothingbutmathproofs71504 ай бұрын
@@PrimeNewtonsperfect response!
@tomctutor4 ай бұрын
pf's you would get: (1/2√x) - (√x/(2(x+6)) don't know about the integration though, maybe you could try that and let us know?
@alifiras11304 ай бұрын
@@tomctutor i will end with ln(∞) and my calculator cant find that value
@tomctutor4 ай бұрын
@@alifiras1130 Ok the integral of the _pf_ form as shown is ∫1/2√x .dx - ∫√x/(2 (x + 6)) .dx = [√x] -[√x - √(6) tan^(-1)(√x/√6)] = √(6) tan^(-1)(√x/√6) then you take _ℓim_ t ->0 part from the _ℓim_ t ->∞ part as in the end of the video to get your answer.
@Bedoroski3 ай бұрын
Anyone figured out how to evaluate this integral by parts? I hardly found any luck