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Current time:0:00Total duration:9:05

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.6 (EK)

let's now explain to ourselves so you could say a more formal communication of the integral test integral test so it tells us that if we assume that we have some f of X if we have some f of X that is positive positive continuous continuous continuous and decreasing and decreasing and decreasing on some interval on so starting at K and including K all the way to infinity then we can make one of two statements we could say either that if the improper integral from K to infinity of f of X DX is convergent is convergent then then the sum the infinite series from N is equal to K to infinity of f of n is also convergent is also convergent convergent and this is actually the case that we saw when we looked at 1 over N squared but I'll look at that in a second but the second claim that we could make or the second deduction that we might be able to make using the integral test is if it's the other way around that if the integral from K to infinity the improper integral of f of X DX is divergent divergent then the same thing is true for the corresponding infinite series then this infinite series right over here is also going to be is also die virgin and as I already mentioned we in the last video we already saw this in the case of f of X is equal to 1 over x squared we saw that since since is integral from 1 to infinity of of 1 over x squared over 1 over x squared DX is convergent in fact it equals 1 it equals 1 because of that we were able we were able to say that the sum from N equals the sum from n is equal to 1 to infinity of 1 over N squared is also convergent also convergent and now we can see an example where we go the other way for example we know that this integral let me write it the integral down let's start with this integral from 1 to infinity knot of f of X is equal to 1 over x squared but let's say that f of X is equal to 1 over X actually let me just write that down let's just start with f of X is equal to 1 over X it definitely meets our conditions that it is positive and let's say we're going to consider it over the interval over the interval from 1 to infinity so it meets this first constraint over this interval 1 over X is positive it is continuous and it is decreasing as x increases f of X decreases so the integral test should apply so let's see what this what the what the improper integral from 1 to infinity of this would be so if we take if we go from 1 to infinity of 1 over X of 1 over X DX this is equal to we could write this as the limit as T approaches infinity of the definite integral from 1 to t of 1 over X DX which is equal to the limit as T approaches infinity of take the antiderivative is going to be the natural log of X the natural log of X going from 1 to T one to T we could do the at well it's really the absolute value of X but we're dealing with positive X's here so it's just going to be the natural log of X which is going to be the limit as T approaches infinity of the natural log of T or I could even say the natural log of the absolute value of T which is just going to be the natural log of T because it's positive T's minus the natural log of 1 minus the natural log of the absolute value of 1 well the natural log of 1 is 0 so it's going to be the natural log of T the limit as that approaches infinity but the limit as that approaches infinity just going to be unbounded this is going to go to infinity this right over here is divergent so this right over here is divergent so this is divergent and because this is divergent we can then say we can then say by the integral test we can then say the integral test once again our function over this interval positive continuous decreasing we saw that this improper integral right over here is divergent and then by the second point of the integral test we can say therefore and I haven't rigorously proved it yet but hopefully I gave you a good intuitive justification in the previous video that the integral R that the the infinite series from N equals 1 to infinity of 1 over N which is the harmonic series that this is also this is also this is also divergent so we've already shown that the harmonic series is divergent using a very beautiful elegant proof by Oh M I think I'm probably mispronouncing his name that use the comparison test but just like this we have used the integral test to show that it is also divergent and once again let's remember what the whole motivation of the integral test is let me draw f of X is equal to 1 over X so f of X is equal to 1 over X would look like my best attempt here so let's say that's 1 two three that is one two and so see when X is one f of X is one when X is two f of X is one-half one-third what's 1/2 here it'll be two over here so it looks like this so this is f of X is equal to 1 over N and once again we see that that over the interval we care about from one to infinity it's definitely positive continuous and decreasing and if we look at this sum right over here we could view this sum as as let's do that let me write it down so the sum the sum from N equals 1 to infinity of 1 over N is equal to 1 plus 1/2 plus 1/3 and of course we keep going on and on and on and on in this case since we want to show us divergent we could say hey look this is an over estimate of of this area here let me be clear so we have this area we have this area in green which is what the improper integral is denoting so that right over there is the improper integral from 1 to infinity of of 1 over X DX now you could view this as an over estimate of that area so this first this one right over here you could say that this is this one height times 1 with so that's that block right over there that's the area of that is going to be equal to 1 then this over here 1/2 you could view that as the area of the next block of the next block and so you can kind of view this as a left-sided Riemann sum I guess is one way to think about it and so this is going to be 1/2 Y a left-sided Riemann sum so this is going to be 1/2 and then the 1/3 is going to be this one it's going to be this one and notice they're all the the actual area we care about that the improper integral it's all contained in these blocks so this is going to be an over is going to be larger than this but we've already seen that this is unbounded towards infinity this is divergent so if this is larger than this and this is divergent this goes to infinity then this must also go to infinity so that's exactly where the integral test is coming from

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