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.
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.
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.
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). 😁
@@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.
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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.
Incredible that a line of infinite length encloses a finite region
Congrats my friend. Like the blackboard and working with chalk. You 're a brilliant example for all the teachers...!!
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.
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.
Isn’t there a little mistake when you put back sqrt(6)? Check the signs. But at the end the result doesn’t change
SQRT( number) is always positive.
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
That is not what they meant...
@@Tomorrow32
Thank you so much ❤
The tumbnail is wrong tho, great video
Thank you. I fixed it
Amazing!!
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). 😁
Simply Great!!!!
Can i solve the integral by using partial fractions?
Try it.
@@PrimeNewtonsperfect response!
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?
@@tomctutor i will end with ln(∞) and my calculator cant find that value
@@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.
Hello, I dont understand why you substituted with root 6 tan rather than just tan theta, why root 6
Im confused how did the inverse tan turn back to tan
Anyone figured out how to evaluate this integral by parts? I hardly found any luck
WoW what a problem