Calculus 2 - Integral Test For Convergence and Divergence of Series

This calculus 2 video tutorial provides a basic introduction into the integral test for convergence and divergence of a series with improper integrals. To perform the integral test, let the sequence be equal to a function that is continuous, positive, and decreasing on the interval [1, infinity). If these conditions are met, then the infinite series will converge if the definite integral from 1 to infinity converges. If the integral test is divergent, then the series is divergent as well. Examples and practice problems include integration techniques such as u-substitution, power rule for integration, trigonometric substitution which leads to inverse tangent functions, and completing the square before integration. The divergent harmonic series is included in this video as well.
Improper Integrals:
• Improper Integrals - C...
Converging & Diverging Sequences:
• Converging and Divergi...
Monotonic & Bounded Sequences:
• Monotonic Sequences an...
Absolute Value Theorem - Sequences:
• Absolute Value Theorem...
Squeeze Theorem - Sequences:
• Squeeze Theorem For Se...
________________________________
Geometric Series & Sequences:
• Geometric Series and G...
Introduction to Series - Convergence:
• Convergence and Diverg...
Divergence Test For Series:
• Divergence Test For Se...
Harmonic Series:
• Harmonic Series
Telescoping Series:
• Telescoping Series
__________________________________
Integral Test For Divergence:
• Calculus 2 - Integral ...
Remainder Estimate - Integral Test:
• Remainder Estimate For...
P-Series:
• P-series
Direct Comparison Test:
• Direct Comparison Test...
Limit Comparison Test:
• Limit Comparison Test
___________________________________
Calculus Final Exam and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
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[@TheOrganicChemistryTutor]

Final Exams and Video Playlists: www.video-tutor.net/ Full-Length Videos & Worksheets: www.patreon.com/MathScienceTutor/collections Next Video: kzbin.info/www/bejne/rJK4Znd6mLmZkM0

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[@akin.kilic.]

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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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[@obww306]

Thanks

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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])

[@RamytRamyt]

You are amazing! Thank you so much!!!!!!

[@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

[@paulmcnamara4962]

You're my favorite person

[@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?

[@ntombenhlejabulile7222]

best lesson ever i'm giong to ace my upcoming tests

[@SirChesterElderberry]

did you ace it?

[@sucukinhiding]

I hope I can someday return to this video and say "I really had to watch this to pass calc 2, now all of my classes are just coding ones.."

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Thanks a lot. From DRC,i have enjoyed it.

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[@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…

[@deanmahori6460]

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

[@PotatoBoiOfficial]

extreme praise for you!!

[@Salamanca-joro]

To the mathematician who invented integral test , I hope you're good in he*ll , bro improper integrals is enough why would you do this again ?

[@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.

[@Epicvampire800]

It depends on your teacher and how pedantic they are. Since he doesnt know who your teacher is he teaches the long route so everyone is covered.

[@jacelemke3006]

@@Epicvampire800 Oh! Nice. I was wondering the same thing

[@fyim6167]

after 6 minutes i finally heard the "you can now use the integral test" kil me -_-

[@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

[@gooz7984]

its mindblowing that a guy on the internet manages to explain things better and quicker than the "top professors" in universities...

[@svgstark3742]

What should we do if the denominator is non factorisable quadratic

[@YangLiu-uh7bp]

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

[@kuroshitsuji2580]

What if we get a negative 0? Is it convergence or divergence? Hope you'll see this. Thank you.

[@RokuSquad]

Negative 0 is 0.

[@beachwave5705]

are u dumb

[@serrairemakman4308]

This is real art of Mathematics

[@jaredkitto9373]

@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

[@basilasar4772]

Nice video

[@rheamiearingo3248]

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

[@andresrobertotulcan6935]

great job

[@pawa241]

Someone tell me if I'm wrong but can't you easily tell if a fractional function is decreasing by simply looking at the exponents on top and bottom and seeing if the order of the bottom is larger than the top? I know why he is using the first derivative test to do the proper solution but I'm sitting here pulling my hair out wondering why he hasn't at least mentioned the shortcut in passing.

[@enriques4nchez]

Thank you!

[@celineal-bukhari4809]

Love how you make things simple

[@diyarkurdi9817]

He absolutely explains everything

[@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?

[@Ashjj1935]

How did du comed 6x ???

[@laminbsanyang2494]

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

[@papertowel818]

Finals in 3 hours

[@hamdial-akkad8339]

THANK U A lot

[@teddy911]

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

[@suneptoshiozukum9034]

Simply the best teacher out there

[@jjy365959]

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

[@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

[@tenalembahiru5336]

sorry mr. 22:00 Minuit example that u did is not correct if u can try to edit that one

[@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

[@Salamanca-joro]

Why tf would we use derivative when it's literally integral test 12:49!??

[@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 ?

[@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

[@nutritionbyelsa]

super epic resource thank you v kindly

[@doublew8815]

27:49 I guess 2tan(inf) should be pi instead of pi/2

[@nyxia8814]

yes ur right, i was thinking the same thing too!

[@_towards_destination]

Thank you sir, you are an inspiration ✨

[@shanem3014]

theres more effeicient ways to tell if the function is decreasing

[@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

[@theanonimo26able]

thannks!

[@JamesBrodski]

Great video! Thank you so much for making it. God bless :)

[@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

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It's high time you changed your KZbin name to MegaMind sir! 🧠🧠

[@Tombalino]

lov u

[@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! :)

[@WoodrowWoodpecker1125]

Any good explanation for LIZARD'S

[@odgarig8601]

A lot of these we don't even need to test if they're decreasing since bottom heavy obviously means decreasing

[@Mirza193]

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[@kidusbahiru4369]

This is wonderful

[@prammar1951]

Last question, it's π you forgot to multiply the 2

[@user-tt3vp1uw6o]

Why does it feel like this video was made in 2007

[@gibbyjibb0]

oh thank god arctan is in here

[@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)

[@nipasy]

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

[@elmo4672]

sign, not sine.

[@Mnzmanzmz]

The final answer you forgot to multiply 2 in pi/2

[@MotivationalDevil-y6k]

Dont forget calculator

[@arafathj2144]

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

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Am I the only one that feels his voice is so satisfying ?

[@JonasAntonioCoquia]

very nice thank you for this video

[@Praveensharma-ns4so]

nice explanation

[@ambersalas4833]

why did he choose tanθ to replace x+3??? I know he said the other form related to arctan but why did he choose that specifically??

[@scarlettq81820]

you could also u sub: u = x+3 and find that integral 2/u^2 + 1 which is in form of antiderivative of 1/u^2 + 1 = tan^-1x so 2tan^-1(x+3)

[@cuzamangelique2673]

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

[@emira1056]

you are the best. thank u

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love you :)

[@m.y.m401]

You really understand it

[@Reactsonshorts786]

So good lectures