Using the Integral Test for Series

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Using the Integral Test for Series - One complete example. I justify EVERYTHING!

[@sl0907]

this guy saved hundreds of thousands of college kids.

[@dreamilyemily]

I will never understand why calculus teachers don't just show your videos in class. My teacher has been trying to explain the integral test for weeks, and nobody has understood it... And you explained it in a span of 8 minutes. Can't thank you enough :)

[@sianmuanthangmilun6104]

Sometimes your voice remarkably changes. :)

[@lifeoftomi_]

Sian Muan Thang Milun the videos are in different years, camera might be newer and/or his voice changed

[@HassaanALal]

he just woke up, i guess.

[@txt0]

That moment when you realise that you have heard Patrick so much that you can tell when he is sick.

[@abdullahbabor4876]

this time his voice dramatically changed!!

[@idontcarewhatname123]

THANK YOU SO MUCH FOR THESE CALC VIDEOS..ive been self studying for the ap calc bc exam and these helped a lot.

[@mikeyura]

I really like your tutorials best for Calc II review on KZbin - they're simple and straight forward. And short but not too short.

[@DallinJackson]

You are brilliant, Patrick! Thank you so much for your dedication to the public's understanding of math!

[@patrickjmt]

@TheMoby37 yes, it diverges as well

[@scotty2hottie469]

That's what she said! 6:10

[@ama004q8]

THAT WAS MY EXAM PROBLEM :@:@:@:@:@!!!

[@patrickjmt]

yes, that is exactly correct! u can not forget about that -1/2 out front. i was helping someone yesterday and kept doing exactly that! i kept forgetting about what i factored out of the integral.

[@patrickjmt]

thanks!

[@jaywalton2944]

Very nice video, everything is very clear and you make a lot of sense. I like that you got rid of the technical jargon and just got down to working the problem. This really helps me learn. Thank you for posting!

[@KurdzOfIslam]

once again you saved my day!!! I have watched other people videos on KZbin for calculus, but none of them explain it and break it down as good as you do Patrick.

[@patrickjmt]

@rickers2011 i didnt mention it? it is like one of the very first things i point out...

[@patrickjmt]

@riliana412 glad i could help you out : )

[@bushammar9]

thank youu.... I was lost in my calc 2 class and now i'm back on track because of people like you who devote their time to upload videos in youtube.

[@patrickjmt]

@dia8449 that is because you do not understand the notion of convergence and divergence i would assume.

[@patrickjmt]

actually, i got a physics book the other day, and it did not look too hard... what sorts of stuff are you thinking about? i plan on learning 1st semester physics

[@johnnymahdessian8673]

You are amazing patrick!! My calc teacher is great but sometimes i missed like 2 days of school and you helped me catch up. For once i dont feel stupid in that class!

[@JDMricist]

WOW! Thanks! So much help! Will be looking at your root test, ratio test, etc... vids next!

[@6ixbertodoh634]

i like how you go into detail makes me more confident

[@Cootiecat07945]

I LOVE YOU. This math problem was basically what was on my homework. I understand convergence/divergence, I was just a little rusty on exponents and logs.

[@markj12]

You are a life saver!!! I'm taking Calc2 online from Iraq. Was lost but now I'm found. Thanks

[@whitlockbr]

Your videos have literally saved my hyde in classes. Thank you. I'd love to see more. I've been watching since calc 1. I have started doing the problems as you write them then seeing if you do them the same way. You and paul dawkins gotta get together and do something with calc and mabye ODE.. that would be cool to see.

[@patrickjmt]

yes, if the function is easy to graph, then go for it. however, on a test / exam, people may want to see you use the derivative so you may ask in that case.

[@edyraper]

0:05 to 0:23 Integral test in only 18 seconds. You are a master. Thank you!

[@patrickjmt]

@agudelostephens i am a bit surprised then... series are used all over the place. you should find some examples so that it might motivate your students a bit more.

[@patrickjmt]

glad to help :)

[@777redoctober]

this was great man! i really appreciate the help

[@patrickjmt]

to mvsparks54: e to a 'large' negative number does get close to zero, but e to a large positive number approaches infinity, not zero.

[@emanuelmcholo4650]

VERY NICE PRESENTATION

[@patrickjmt]

@vivalalaa uh... okie dokie.

[@chitterchatter75]

I don't even go to class anymore. I just watch your videos, show up for the tests, and do better than everyone else. Thank you so much for being born.

[@sonicspeedfan1]

honestly, you are a lifesaver for first year students at uni. I can not thank you enough for all your hard work!

[@patrickjmt]

yep, that is part of the challenge

[@patrickjmt]

glad it helps... i just dont like the blue though

[@yiwawww]

I love your videos! They help a lot!!!!!!!!!

[@patrickjmt]

@jeohwai if you are doing ap math in high school, you can not suck at it that much : )

[@andiibart]

This is a great test review!

[@jescudi_]

When Factoring out the -1/2 im the beginning steps of finding the limit. It should have became e^-t^2 + e^-1^2. Not e^-t^2 - e^-1^2. Also, e^-1^2 is just equal to e, not e^-1. Meaning the integral converged to e. It doesn't make any difference with your final answer but I just thought I'd point that out.

[@patrickjmt]

glad i could help out ;)

[@iurysza]

Dude, you rock!

[@ian559fresno]

Thanks

[@maddymc2017]

I LOVE YOU. If I pass my calc final I will literally buy you a dozen roses because you are freaking great. Thank you omg.

[@FarFromEquilibrium]

thanks this was very helpful. it used a lot of different methods, thats tedious but helpful.

[@derp6632]

Do you have a video with an example where there is a variable within the series and the solution to the series is given and you have to solve for that variable? Also, do you have a video where you have to solve for the x values within a series that make the serious convergent(2 values), and then to find the sum of the series within those convergent values?

[@queenb1292]

Wow thank you!

[@mishnareemincey]

Half the video was on figuring out if you could even use the freakin Integral Test lol

[@kevsac94]

wow this helped so much.thanks!

[@Arianna0010]

thanks! your videos are excellent! congratulation ;) i really like them, because i'm studying mechanical engineering here in venezuela and your videos had helped me a lot when i need to understand how to do some exercise!! i think that you should make more videos with difficult problems! that would be great! ;)

[@patrickjmt]

original variable = original limits of integration. it is correct!

[@2014xckt]

this is the exact problem I was just working on. you're amazing

[@amineawad]

THANKYOU SIR!

[@patrickjmt]

calculus teacher of the world!

[@BIGHARV29]

"we're only going from 1 to infinity, anyway..." LOL love your vids man, 3rd year uni engineering and still watching them! great refreshers and more examples than one could ever need. In due time, you will get much recognition for your hard work... power to the nerds dude!!

[@angrydango]

I'm just gonna say, whoever gave you a thumbs down just doesn't know ANYthing about math because you make it so incredibly easy to understand. Like someone else said below me, I stopped going to class because your lectures are not only more simple but you get the point across in way less time than my professor. Thanks so much for sharing your knowledge!

[@MrBlash93]

Amazing. Thx u

[@aepassion]

@PythagosaurusRex The concavity test is done via the second derivative test. He used the fist derivative test, instead.

[@lopez5026]

@dia8449 look at the graph to help you. Convergence means that the graph is approaching some number. Divergence means that the graph is infinitely growing aka infinity.

[@patrickjmt]

happy to help! : ) just send me part of your tuition!

[@GohansRiceBall]

Thank you.

[@ENr369]

YOU ARE AWESOME!

[@pablogonzaleznegredo5853]

And what are we computing when finding the improper integral? Is that -1/2e something?

[@12TSD]

great video really helped Just to follow up on that same question how would you determine how many terms are needed to obtain an approximation accurate to within 0.000001?

[@saifehsan]

i want you in my university now !!! man you are freakin' awesome i can't believe ,you make it that easy !!! thanks dude you saved me

[@eparra0506]

When would you use this test?

[@justrandom_256]

thanks. u are good

[@rickers2011]

@ehabq yes, you would still get a negative number if you put 1. The function must be positive and decreasing from 1 to infinity--he didn't mention this part--for the integral test to be applicable.

[@kathryniii]

So what if the series was not positive or decreasing after all the checks and that means integral test fails. In this case, what other test can I use to see if the series is Conv or Div?

[@Alainabh]

you have saved my life, no lie.

[@YairCat]

You are good, i cannot argue with that. But i do not understand how can we get the number it converts to

[@memaaa910]

@Syed Faraz if one of them divergent the final is also divergent :)

[@kylewang2160]

Doesn't the function need to be continuous too,in order to apply the integral test?

[@Svenskivsk]

I do have a some questions. How do you calculate the value to which the series converges? Is the value of (1/2e) significant other than showing the integral and therefore the series are convergent?

[@JesssMooo]

Wouldn't the limits of integration have to be changed to compensate for the u-substitution that was used? Thus making a= -1 & b= (infinity). u(1)= -(1)^2 --> -1 u(infinity)= -(infinity)^2 --> Infinity Or are we assuming that the entire quantity is being squared? (-x)^2

[@purplecow265]

am i right by saying, the series n*e^-n^2 converges because the integral of x*e^-x^2 is 1/2e, which is a finite number? thanks

[@lindap2012]

ooo i just reviewed the integral test today for my final, and this was the same problem i did! haha, oh well more review

[@ahmadbelhaj1756]

where 2x^2 come from when you factored (e^(-x)^2 - 2xe^(-x)^2 ===> e^(-x)^2(1-2x^2)

[@LizardanNet]

what happened to the negative sign of the formula at the end? where you had " -e^-1 "?

[@jazzmastr65]

@JesssMooo No, because he returns the integral to it's variable

[@thayssasouto6101]

I love the videos, and besides learning calculus, also have to learn English because I am Brazilian and sometimes get lost in the details. =]

[@kphamu]

@JaktheAtheist, what book are you using that gives you simple problems? That exact problem was on my exam. I'm thankful that Patrick uses "hard" examples. So hush up about wanting simple problems, this is college level math. "e" is everywhere.

[@DanceInstrumentalist]

Can you solve this using the p-series test, because if you rewrite it as n/e^-n^2 could you say that n is the power and it cannot equal 0, so then p would be greater than 1 and converge?

[@marelynrios3889]

Shouldn't f'(x)=e^-x^2(2x^2+1) since the derivative of e^-x^2=-e^-x^2*-2x?

[@MrDewloo]

***** explain?

[@jescudi_]

Although that is correct, you still have to apply the chain rule because of the other variable x in front

[@nickpollifrone2979]

When you factor out the e^-xsquared when taking the derivative to prove it's decreasing, why do you get a 2x squared instead of just the 2x that's in the previous step

[@renirasa]

+Nick Pollifrone because its in the form of e^(-2x) + x * [ e^(-2x) * 2x ] , x is going to multiply both e^(-2x) and 2x, with, that is what gives the 2x^2

[@AnhPeaches]

I noticed that the answer you evaluated is 1/2e. But when I did it, I got e/2. You plugged in 1 for x but I got e^-1^2, which equals e^1 (you got e^-1). Unless what you mean is that it is suppose to be e^-(x^2).

[@MadJDMTurboBoost]

4:03 less than one*

[@captainobvious90]

Why do we need to find the critical numbers for the f'(x)? Can we just show that f'(x) is decreasing for x larger and equal to one?

[@TheSupertramp11]

Patrick, f ' (x)= e^-x2 (1 - 2x) , you have found the critical numbers with (1 -2x^2)=0 instead.of (1 - 2x)=0 other than that, thumbs up man thanks for your videos,

[@Googs5]

Why at the end doesn't e^-t^2 become a positive infinity because you are squaring it? Same for e^-1^2 where the negative one is squared but you seem to skip that part.

[@TheKamran46]

i think he forgot to square the 1. answer should be e/2

[@rdvonz]

It would be -1/2(-e^(-1)) Distribute that out, negative multiplied by a negative is positive.

[@eracela8341]

Help: Divergent Test says, if lim n to infinity (An) is not equal zero then it is divergent. Kinda confused at the ending. You got 1/(2e) for lim t to infinity 1/(e^t^2) and it converges. Shouldn't it diverge? Please clarify

[@jasonkares3705]

if you have e^(-x)^2, when you plug in the limit of 1, wouldn't that make it just e? like this, e^(-x)^2 e^(-1)^2 e^1 e then youre final answer would be e/2? Not that it makes a difference in convergence, i was just curious about the algebra.

[@YawningBoy]

@patrickJMT haha yeah i know i'm rusty on some of my math before calculus. i looked it up i remember it now.

[@abbykimz]

my question is I/xlogx, where x E [3 to infinity], i get [x] where inifinity is still the upper limit and ln 3 becomes the lower limit, but my series while tend to infinity which means it diverges, did i get the question right?

[@kevsac94]

am i able to graph a function and tell whether it will decrease or increase instead of doing that whole f(x)' ?

[@SyedFaraz]

patrick, if one limit is divergent and other is convergent so whts the final limit ?

[@nataliatorres298]

Am I not supposed to do the first derivative test first to see if it is decreasing, in order to do the integral test? I believe it is crescent...

[@nelmsters]

if you don't want to use this method, then is there another test you can use for this example?

[@patrickjmt]

@xDAZZE i am always here...