Using the Integral Test for Series
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- Опубліковано 10 жов 2008
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Using the Integral Test for Series - One complete example. I justify EVERYTHING!
this guy saved hundreds of thousands of college kids.
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 :)
Sometimes your voice remarkably changes. :)
Sian Muan Thang Milun the videos are in different years, camera might be newer and/or his voice changed
he just woke up, i guess.
That moment when you realise that you have heard Patrick so much that you can tell when he is sick.
this time his voice dramatically changed!!
THAT WAS MY EXAM PROBLEM :@:@:@:@:@!!!
That's what she said! 6:10
I really like your tutorials best for Calc II review on UA-cam - they're simple and straight forward. And short but not too short.
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.
You are brilliant, Patrick! Thank you so much for your dedication to the public's understanding of math!
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.
THANK YOU SO MUCH FOR THESE CALC VIDEOS..ive been self studying for the ap calc bc exam and these helped a lot.
once again you saved my day!!! I have watched other people videos on UA-cam for calculus, but none of them explain it and break it down as good as you do Patrick.
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.
0:05 to 0:23 Integral test in only 18 seconds. You are a master. Thank you!
Half the video was on figuring out if you could even use the freakin Integral Test lol
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.
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.
@TheMoby37 yes, it diverges as well
honestly, you are a lifesaver for first year students at uni. I can not thank you enough for all your hard work!
You are a life saver!!! I'm taking Calc2 online from Iraq. Was lost but now I'm found.
Thanks
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!
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!
@dia8449 that is because you do not understand the notion of convergence and divergence i would assume.
i like how you go into detail makes me more confident
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.
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.
@jeohwai if you are doing ap math in high school, you can not suck at it that much : )
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.
@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.
to mvsparks54:
e to a 'large' negative number does get close to zero, but e to a large positive number approaches infinity, not zero.
WOW! Thanks! So much help! Will be looking at your root test, ratio test, etc... vids next!
"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!!
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
thanks!
@rickers2011 i didnt mention it? it is like one of the very first things i point out...
this was great man! i really appreciate the help
this is the exact problem I was just working on. you're amazing
@riliana412 glad i could help you out : )
calculus teacher of the world!
VERY NICE PRESENTATION
original variable = original limits of integration. it is correct!
I love your videos! They help a lot!!!!!!!!!
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! ;)
This is a great test review!
Dude, you rock!
thanks this was very helpful. it used a lot of different methods, thats tedious but helpful.
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
glad it helps... i just dont like the blue though
yep, that is part of the challenge
glad to help :)
wow this helped so much.thanks!
YOU ARE AWESOME!
@PythagosaurusRex The concavity test is done via the second derivative test. He used the fist derivative test, instead.
Wow thank you!
you have saved my life, no lie.
THANKYOU SIR!
glad i could help out ;)
@vivalalaa uh... okie dokie.
this was the EXACT question i got on my test!! i got 7/8 points on this problem. soo glad i took a look at this video first
glad you like my stuff :) you know where to come to first!
@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.
happy to help! : )
just send me part of your tuition!
yet another amazing video :')
where 2x^2 come from when you factored (e^(-x)^2 - 2xe^(-x)^2 ===> e^(-x)^2(1-2x^2)
@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.
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).
you say: e to anything isn't a positive number
i say: sure it is
you say: e to a negative would be getting closer and closer to zero. which wouldn't be greater than zero.
i say: well it is close to zero, but not zero, and it is positive, so yes, it is greater than zero.
Without you I would've dropped my math class already.
Amazing. Thx u
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?
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!
Thank you.
Thanks
That was beautiful!!! And I was Here trying to read my stupid Math Text Book...Thank You So Much!!!!! I Hope Ace My Exam
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,
ooo i just reviewed the integral test today for my final, and this was the same problem i did! haha, oh well more review
It would be -1/2(-e^(-1)) Distribute that out, negative multiplied by a negative is positive.
I love the videos, and besides learning calculus, also have to learn English because I am Brazilian and sometimes get lost in the details. =]
Thank you SOO MUCH!!!
-a math 222 student
God I love you, thank you for all these videos.
@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.
thanks. u are good
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?
And what are we computing when finding the improper integral? Is that -1/2e something?
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?
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
When would you use this test?
Note that the negative is not in parentheses. so it would go like this
e^-(x)^2
e^-(1)^2
e^-(1)
e^-1
1/e
And then multiple the 1/2 to get..
1/2e
Hey Patrick, you could've just rewrote the function by writing it as a fraction and state that e^x^2 is greater than x for all x. Which means that its decreasing for all x. It would've saved you video time. Cheers.
@patrickJMT, I'm a little confused, sir. When you factored out e to the -x^2 power, I do not see how you got (1-2x^2). Shouldn't it be just (1-2x)?
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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
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?
Doesn't the derivative of e^-x^2 = (-2x)*e-x^2? Great video by the way, not only am I not going to fail Calc II this summer but I am getting an A. Subscribed!
good shit
@xDAZZE i am always here...
am i able to graph a function and tell whether it will decrease or increase instead of doing that whole f(x)' ?
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
what happened to the negative sign of the formula at the end? where you had " -e^-1 "?
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?