improper integrals Types 1 and 2

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improper integrals Types 1 and 2

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Prime Newtons

4 ай бұрын

In this video, I showed how to rewrite and compute an improper integral of both types.

Пікірлер: 25

@user-xw6ky8ob4l 4 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.

@chengkaigoh5101 4 ай бұрын

Incredible that a line of infinite length encloses a finite region

@cadenpink316 4 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-ht7ug 4 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

@Tomorrow32 4 ай бұрын

SQRT( number) is always positive.

@AquaticWaters 4 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

@joaomane4831 Ай бұрын

That is not what they meant... @@Tomorrow32

@glorrin 4 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.

@antonionavarro1000 4 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.

@wolfwittevrongel8067 4 ай бұрын

The tumbnail is wrong tho, great video

@PrimeNewtons 4 ай бұрын

Thank you. I fixed it

@Annihilator-01 4 ай бұрын

Thank you so much ❤

@iquesillos12 4 ай бұрын

Amazing!!

@tomctutor 4 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). 😁

@samueldarko9360 13 күн бұрын

Hello, I dont understand why you substituted with root 6 tan rather than just tan theta, why root 6

@saarike 3 ай бұрын

Simply Great!!!!

@alifiras1130 4 ай бұрын

Can i solve the integral by using partial fractions?

@PrimeNewtons 4 ай бұрын

Try it.

@nothingbutmathproofs7150 4 ай бұрын

@@PrimeNewtonsperfect response!

@tomctutor 4 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?

@alifiras1130 4 ай бұрын

@@tomctutor i will end with ln(∞) and my calculator cant find that value

@tomctutor 4 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.