Calculus 2 - Integral Test For Convergence and Divergence of Series

[@TheOrganicChemistryTutor]

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[@theobserver314]

For the last problem, it was supposed to be π-2arctan(4).

[@tsuulifts]

correct, because of the factor of 2 infront of the arctan(inf)

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[@georgesadler7830]

Professor Organic Chemistry Tutor, thank you for another fantastic video/lecture on the Integral Test for Convergence and Divergence of Series in Calculus Two. This part of Calculus Two can be problematic, however students must do many problems in order to fully understand this material. This is an error free video/lecture on KZbin TV with the Organic Chemistry Tutor.

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[@hsc4648]

27:48 shouldn't the 2 in front cancel out the 2 from pi/2? Final answer should therefore be pi-2 arctan(4)?

[@ALSAYED-yg5jp]

Yeah I noted that too, I don't know I think he mis that step

[@nelsonasemota7994]

Yea I was just thinking that

[@jimb4549]

I just checked with an integral calculator and he made a mistake, the final answer should be [π-2tan⁻¹(4)]

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[@dillpickleboy1128]

23:39 if u dont understand the trig sub: if a function looks like (1/[(a^2)+(u^2)]) where a^2 is some constant term and u is some variable arctan formula takes the form 1/a * arctan(u/a) for this problem. take the two in the numerator out of the integral replace (x+3) with u a = 1 2 * (1/1 * arctan[(x+3)/1])

[@mr.9754]

12:20 Don't waste time "testing" the regions. Just quickly graph (since the denominator is always positive) y=8-6x^2 which is easy since u have the roots (and u understand end behavior) and you'll get a parabola that opens down.

[@thedarkdev3439]

Most college calculus courses don’t allow you to use a calculator for quizzes and exams. This is a hand drawn method for those who can’t use a calculator or understand the methods behind it. But I totally get your point

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[@obww306]

Thanks

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He's got a Patreon linked in the description

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[@fibrealex]

For the series 2n/3n^2 + 4: Why do the first derivative test to check if it's decreasing? Wouldn't you just be able to check if it was decreasing by seeing that the power on the bottom is greater than the power on the top? Or, if you just plugged in the first and second terms. s1 = 2/7 > s2 = 1/4, therefore its decreasing. If the integral from 1 → infinity of 2x/3x^2+4 was evaluated, you could do u-sub and you'd get the same answer but with way less work.

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[@ti84satact12]

If you can graph the function and can see that it decreases will that suffice to show that the function decreases or must you take the first derivative?

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[@JonasAntonioCoquia]

very nice thank you for this video

[@qixuanhe5656]

For the last example, is the final answer not PI minus 2 times arctangent of 4? Since you have the constant 2 in front of the whole thing

[@svgstark3742]

What should we do if the denominator is non factorisable quadratic

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[@Young-keet]

@13:43, how do you know that it is continous at [2, infinity)?

[@dillinsingh5887]

for every input value on interval [1, inf ) there is always an output

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[@teddy911]

the last limit is supposed to be π-2acrtan(4)

[@lifefan1]

Thanks as always 😊

[@nathanmanning4961]

I believe that you could have just done a u-sub instead of a trig sub for the last problem, would have given you 1/(u^2 + 1) which is the derivative of arctan, plug the u back in for your integral. Great video though.

[@almscurium]

Arctan is like the definition of a trig sub…

[@PaulOtoo-n3e]

Please how do you get to know the genre of test you're to go about or use if the question doesn't state it ?

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[@m.y.m401]

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[@jjy365959]

at 27:54 did the tutor for got to put 2 on pi/2? because it has 2 in than^-1 (b+3)

[@Tonydalowk]

Looks like he did forget

[@lilgtBcG]

how do you know whether to use the p-series or integral test?

[@YangLiu-uh7bp]

I am still confused about replace the (x+3)^2 with tan theta

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[@ezilemdodana3969]

Why are you not changing the limits of integration after making the u-substitution?

[@fibrealex]

as long as you sub in for u at the end, you're changing the integral back to terms of x, so you can use the same bounds for x when evaluating that integral

[@diags_1109]

at 24:00 why did you not just use u sub for x+3 instead of trig sub, is it because how profs teach it or something because 1/u+1 du is inverse trig so it would be 2tan^-1(u) and u is x+3

[@nyesigadaniel7331]

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[@zainnabeel1780]

I have a question if the integral converges to an N value does the series also converge to that N value or do we just know that it converges ?

[@allisonjiang2703]

They don’t converge to the same value

[@enriques4nchez]

Thank you!

[@deanmahori6460]

Please help Summation ln(n/n+1)) determine if it converges

[@rheamiearingo3248]

In the second example, why does du becomes 6x dx? Should it be just 2x dx?

[@mbaziirafaysal]

How do you know it is positive, continuous and decreasing

[@hamdial-akkad8339]

THANK U A lot

[@robertolopez6527]

Watching this before exam

[@WoodrowWoodpecker1125]

Any good explanation for LIZARD'S

[@gamingforever3170]

best for preparation

[@nononnomonohjghdgdshrsrhsjgd]

why is not necesssary to find the limit of convergence of the series? we prove with the test that there is a converg. or divergence. how to find the convergence of the series?

[@becausewhynot7183]

uh question: Didn't the 2's cancel each other out at the end? Thank you for going into detail btw!

[@korra5815]

I guess they did, was looking for a comment if anyone else has realized

[@sarkersaadahmed]

13:29 could you / anyone explain why we have to now focus from 2 to infinite

[@obww306]

We have a condition that our function has to be decreasing on the intervals [1, infinity). But we saw that it is not. Instead of this we took from 2 to infinity because it is decreasing in this interval. (im not sure about it 100%, and sorry for my english)

[@arafathj2144]

What happen if n starts from 2 Does it satisfy the integral test

[@laminbsanyang2494]

Sir what if you take the interval of the 2nd Question from 1 to infinity

[@nipasy]

12:02 Is he saying "sine chart"? (That's what the auto-generated subtitles say)

[@elmo4672]

sign, not sine.

[@samkyoung]

why didnt he change the bounds when he did u-substitution, don't you have to since it is a definite integral?

[@YousufZamanOfSteel]

You don't have to change the bounds as long as you substitute x back for u before you evaluate

[@samkyoung]

Yousuf Zaman thank you so much

[@YousufZamanOfSteel]

@@samkyoung np!

[@mehmetbilgili4155]

27:32 Shouldn’t it have to be 2.(pi/2)?

[@baharsalman3439]

I thought the same

[@davidvilla1247]

Is this really calc 2? I am taking calc 3 right now and just learning this

[@logicbewithyou111]

yes. im taking cal 2 rn and im learning this.

[@davidvilla1247]

@@logicbewithyou111 oof I have a calc 3 test on this lol but after I this I think I will learn power series

[@connormazzaferro6213]

if the value of n under the sigma was not 1, would the range of continuity change to whatever that value is too infinity ? for example if n=2, would the range that has to be continuous change to [2, infinity)?

[@ivyreed2385]

Yes, you could simply evaluate from 1 to infinity, then subtract the term at n=1 if you are evaluating from n=2 to infinity (take out that partial sum since it is not included in the interval). If n equals a larger number, you would apply the same principle and subtract however many partial sums there are that aren't included in the interval. Hope this helps! :)

[@cuzamangelique2673]

if the resulting integral value is zero in an integral test, does the series converges?

[@vatra6799]

so many calculus 1 wanna cry 🙂

[@weaver2609]

when you do u-sub, dont you have to make your new points on the definite integral?

[@mikuculus3720]

Optional

[@claude7668]

pls do we know why he chose to replace x with b, just curious. It seems that x will still be fine to me

[@kholoudhakim7887]

he is using improper integration and b refers to the larger number which here is infinity

[@guninkakar7714]

love you :)