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Current time:0:00Total duration:2:42

AP.CALC:

LIM‑6 (EU)

, LIM‑6.A (LO)

, LIM‑6.A.1 (EK)

, LIM‑6.A.2 (EK)

right here I've graphed part of the graph of y is equal to one over X and what I'm curious about is the area under this curve and above the x-axis between x equals one and infinity so I want to figure out this area right over here so let's try to do it so we could set this up as an improper integral going from 1 to infinity of 1 over X 1 over X DX well we can once again we can view this as the limit the limit actually me do that same yellow color I like that more we can view this as the limit as n approaches infinity of the integral from 1 to N of 1 over X DX which we can write as the limit as n approaches infinity of the antiderivative of 1 over X which is the natural log of the absolute value of x so this is going to be the natural log of the absolute value of x and the absolute value of x won't really matter so much we could just say X because we're dealing with positive values of x but I'll just write down as the natural log of the absolute value of x between X is 1 and X is N and so this is going to be equal to the limit as n approaches infinity of you evaluate this at n so you're going to get the natural log I could write the absolute value of n but we know that n is going to be positive so we can just write the natural log of n minus the natural log of the absolute value of 1 or the natural log of 1 natural log of 1 is just 0 e to the 0 power is equal to 1 so this boils down to the limit as n approaches infinity of the natural log of n now this is interesting natural log function just keeps getting larger and larger and larger it kind of natural log function looks something like keeps growing and growing and growing like this I'll be at a at a slower and slower pace but it keeps growing the limit as n approaches infinity of the natural log of n is just equal to infinity so here we do not have a finite area this is an infinite this is an infinite area it's interesting when this function decreased faster when it was 1 over x squared we had a finite area now we have an infinite area and so we would say that this integral right over here is this improper integral is divergent divergent and we're done

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