The Integral Test - Proof

[@rodioniskhakov905]

Man, that was a crystal clear explanation. My respect!

[@minus4025]

Very clear and easy to follow. Thanks.

[@sumsumiho]

Thank you so much for such a detailed and easy-to-follow explanation!

[@kaleyschuster1951]

Thank you for walking through this proof in such a clear, direct way. I am taking Analysis and this helped me a lot

[@raghumalhotra3490]

Lovely bro...God bless you

[@muhammedalshaer3333]

Great content!👍

[@amoghviswanath8264]

really love ur videos man, thank you so much for your help!!!

[@DenisBencic]

This was very comprehensive. Thank you :)

[@MuhammadSaqib]

Hats offf

[@joehd2009]

Amazing thank you

[@jamesrobertson9149]

very good.

[@diamondmat4380]

thx !it help me a lot to get it

[@danielmichaeli2633]

Thanks for the intuitive explanation! One question though: why do we use the ≤ ≥ signs instead of ? Is it not obvious that the area of the rectangles will in fact be larger / smaller than the area under the curve? In what case will they be equal?

[@slcmathpc]

You can imagine an unusual function where parts of it are constant over some intervals.

[@shreyjoshi4891]

Thank you for the clear and concise explanation. I spent a while trying to understand the wikipedia proof for this but it was too verbose for a beginner.

[@slcmathpc]

That is the downside of Wikipedia; it is a great repository of information but not always presented in an intuitive fashion.

[@AB-tl5xb]

thnx a lot :)

[@harshdmp0]

Thanx sir :)

[@sunitapandey1195]

Aap DTU me mechanical engineering ke student jo na

[@mktbhandari]

Did you not miss that the function f should be continuous?

[@slcmathpc]

It is enough for the function to be Riemann integrable, which is a weaker condition than continuity, but this gets a bit more technical so this is why I omitted it.

[@Rsingh1]

Can i know the book from which its taken ?please!!

[@slcmathpc]

I do not use a book, sorry. I simply try to present ideas in the clearest and most intuitive fashion.

[@Rsingh1]

@@slcmathpc oh okay

[@michaelqi8443]

why does the function need to be eventually decreasing? can it be increasing?

[@slcmathpc]

To make the desired inequalities work, we need the function to be eventually decreasing. Moreover, if the function is increasing, then the individual terms of the series will not converge to zero and so the series will diverge by the Divergence Test yielding a much simpler problem.

[@michaelqi8443]

+slcmath@pc but if its increasing, dont the inequality signs just flip? then unless im missing something, that would allow the proof to be made regardless if it is decreasing. for example, 1/n diverges from the integral test but why do you have to prove its decreasing first? if you take the integral immediately you still end up with its divergence.

[@slcmathpc]

Yes, you can make it work but when the function is increasing, the much simpler Divergence Test applies to show divergence of the series; it's all in the spirit of keeping things as simple as possible. :-)

[@michaelqi8443]

slcmath@pc then you wouldnt have to prove its decreasing, would you?

[@slcmathpc]

Not every function is eventually increasing or decreasing; the question must be investigated.

[@JoffreyB]

Why u haven’t taken in the second picture integral up until N+1? I mean u could summarize with right rectangles as well as with same previous points. Why did u pick up just until N?

[@arsenron]

Because he takes right boundary of rectangles: when you reach N, your last rectangle will have N as its right boundary. Vice versa, in the first case, when you reach N, you will have N as left boundary and N + 1 as its right boundary respectively.

[@JoffreyB]

i can't understand how did u merge two COMPLETELY DIFFERENT series, which represent DIFFERENT SUM into one inequality?? Somebody please explain this to me

[@D3tyHuff]

They aren't different series. The first one is the function evaluated on the left end of the sum of rectangles, while the other one is the function evaluated on the right hand of the rectangles, and both summed up to N. If you evaluate the right hand side of the rectangles, all you really do is start at n=2, therefore when you add the first term (a1), you get the summation of the left hand side of the rectangles (starting from n=1 to N), and you can bring the two integrals into an inequality.