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...
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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
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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
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i appreciate that he doesn't skip steps because he thinks its "easy" or everyone should know how to do that step. I have a lot of smart professors who think people are as smart as them in a subject and fail to go over all the steps, which makes it harder for everyone to keep up
My professor is like this and I think that's the main reason I don't understand his classes while I am perfectly fine over here, some student had the courage to ask how did he get that result because he skipped like 3 steps and he ridiculed the student in front of everybody for not understanding the steps he skipped because we should have learnt all of this perfectly in highschool...
I know that feel mate...
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My professor while a smart guy and a good teacher, he will constantly go over steps and it sometimes takes me a couple moments to figure out what shortcut he took. While I'm doing that he is already moving on with the material and it can be annoying.
He’s got this “Dora the Explorer” pace where he wants you to answer yourself lol
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For the last problem, it was supposed to be π-2arctan(4).
correct, because of the factor of 2 infront of the arctan(inf)
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the last limit is supposed to be π-2acrtan(4)
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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])
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 KZhead TV with the Organic Chemistry Tutor.
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27:48 shouldn't the 2 in front cancel out the 2 from pi/2? Final answer should therefore be pi-2 arctan(4)?
Yeah I noted that too, I don't know I think he mis that step
Yea I was just thinking that
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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there is so much information to absorb in calculus. he always goes over simple things that i have already forgotten ....
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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.
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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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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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.
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.
@@Epicvampire800 Oh! Nice. I was wondering the same thing
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
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Sir what if you take the interval of the 2nd Question from 1 to infinity
In the second example, why does du becomes 6x dx? Should it be just 2x dx?
What should we do if the denominator is non factorisable quadratic
you are the reason i am passing Thank you
at 27:54 did the tutor for got to put 2 on pi/2? because it has 2 in than^-1 (b+3)
Please help Summation ln(n/n+1)) determine if it converges
What happen if n starts from 2 Does it satisfy the integral test
I am still confused about replace the (x+3)^2 with tan theta
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.
Arctan is like the definition of a trig sub…
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Excellent
You really understand it
if the resulting integral value is zero in an integral test, does the series converges?
Barakalla 🤲 jazakkalla
Nice video
@13:43, how do you know that it is continous at [2, infinity)?
for every input value on interval [1, inf ) there is always an output
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 ?
They don’t converge to the same value
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)?
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! :)
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
Any good explanation for LIZARD'S
thannks!
pls do we know why he chose to replace x with b, just curious. It seems that x will still be fine to me
he is using improper integration and b refers to the larger number which here is infinity
Am I the only one that feels his voice is so satisfying ?
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when you do u-sub, dont you have to make your new points on the definite integral?
Optional
uh question: Didn't the 2's cancel each other out at the end? Thank you for going into detail btw!
I guess they did, was looking for a comment if anyone else has realized
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?
Why are you not changing the limits of integration after making the u-substitution?
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
love you :)
13:29 could you / anyone explain why we have to now focus from 2 to infinite
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)
What if we get a negative 0? Is it convergence or divergence? Hope you'll see this. Thank you.
Negative 0 is 0.
are u dumb
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.
any one know about the integral test applications
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??
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)
How did du comed 6x ???
Could someone explain why we differentiate 2x/(3x^2 + 4), instead of just integrating with the given function and solving for it at infinity and 1? Basically, why is the second problem solved differently than the previous one? Thank you!
We differentiated so we could use the first derivative test to see if the function decreases on [1,infinity)
12:02 Is he saying "sine chart"? (That's what the auto-generated subtitles say)
sign, not sine.
27:32 Shouldn’t it have to be 2.(pi/2)?
I thought the same
saved once again..
why didnt he change the bounds when he did u-substitution, don't you have to since it is a definite integral?
You don't have to change the bounds as long as you substitute x back for u before you evaluate
Yousuf Zaman thank you so much
@@samkyoung np!