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AP Calc: LIM‑6 (EU), LIM‑6.A (LO), LIM‑6.A.1 (EK), LIM‑6.A.2 (EK)

Right here I have graphed
part of the graph of y is equal to 1/x. And what I'm curious about
is the area under this curve and above the x-axis between
x equals 1 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/x dx. Well once again-- actually, let
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/x dx, which we can
write as the limit as n approaches infinity of
the antiderivative of 1/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 it
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 could 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 a 0. e to the 0-th 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. The natural log function keeps
growing and growing and growing like this, albeit 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, this improper
integral, is divergent. And we're done.

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