WEBVTT
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PROFESSOR: Hi.
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Welcome back to recitation.
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In lecture you've been
doing definite integration
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for the last couple of lectures.
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And you've just started
with the fundamental theorem
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of calculus.
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So, that gives us this
whole powerful tool
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to compute a bunch of integrals.
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So, with that in mind, I've
got a couple questions for you.
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So compute the
definite integrals.
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The integral from pi over
6 to pi over 3 of tan x dx.
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And the integral of minus pi
over 3 to pi over 3 of tan x
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dx.
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So why don't you
pause the video,
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take a couple minutes to
work on those, come back
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and we'll work on them together.
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All right.
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Welcome back.
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So, as we were saying
we have a couple
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of definite integrals here.
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So one tool for definite
integrals is Riemann sums.
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So, in principle, we could write
down a Riemann sum for, say,
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this first integral here.
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And then try and compute it
by trying to compute the sum
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and taking a limit of it.
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But I think it's pretty clear
that, for a function like tan
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x, that's going to be
really, really hard to do.
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So we want to not use
Riemann sums here.
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And the way we can
not use Riemann sums
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is using this great
tool that we have,
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which is the fundamental
theorem of calculus.
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So the fundamental
theorem of calculus says,
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when you have a definite
integral, instead of computing
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Riemann sums, what you can do
is compute an antiderivative
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and use the antiderivative to
find out what the integral is.
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Just compare two values
of the antiderivative
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and that'll give you the value
of the definite integral.
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So, in particular, let's do
this first integral first.
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So we have the integral from
pi over 6 to pi over 3 of tan
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x dx.
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Well, so, we know what the
antiderivative of tan x is.
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We did that in an
earlier recitation.
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So that's ln of cosine x.
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Or ln of the absolute
value of cosine x.
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So, the fundamental
theorem of calculus
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says that the value of
this definite integral,
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in order to compute
it, we just take
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the difference of that
antiderivative at pi over 3
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and at pi over 6.
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So, by the fundamental
theorem of calculus
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this is equal to ln of the
absolute value of cosine
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x for x between pi
over 6 and pi over 3.
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So, OK, so this, you know,
this is this notation
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that Professor
Jerison introduced.
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So what does this actually mean?
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It's just a shorthand for ln
of the absolute value of cosine
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of pi over 3 minus ln
of the absolute value
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of cosine of pi over 6.
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Now, we could leave
our answer like this
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but we can also
manipulate it and put it
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in a little nicer form.
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So, cosine of pi over 3 is-- oh,
I'm sorry I've made a mistake.
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So it's not ln of
absolute value cosine x.
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It's minus ln of absolute
value of cosine x.
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Sorry about that.
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So this is minus ln cosine
pi over 3 plus-- OK, so
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minus and minus there is plus.
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All right.
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Sorry about that.
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OK, so, good.
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All right.
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And we can pick up where I was
in the middle of my sentence
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and OK, so we can just evaluate
this by just evaluating.
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So cosine of pi over 3,
that's going to be 1/2.
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So this is minus ln and absolute
value of 1/2 is just a 1/2.
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Plus cosine of pi over 6
is square root of 3 over 2.
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So plus ln square
root of 3 over 2.
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And now you can use
your logarithm rules
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to combine those into
a single expression.
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So this is equal to ln
of square root of 3.
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Or you could even use one more
logarithm rule if you wanted.
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So this is equal to 1/2 ln of 3.
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So all of those are, you
know, equivalent expressions
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for the same number,
which is the value
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of this definite integral.
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OK, so, for the second
integral now, we
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can do exactly the same
thing if we wanted to.
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And it'll be a very
similar process.
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You'll write down this
antiderivative here,
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you'll take the
difference in values.
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We can also do something a
little bit clever for this one.
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Which is, definite
integrals have
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some geometric
interpretation, right?
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And if you think
about the function tan
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x-- so we could put up
a little graph here--
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so in between minus pi
over 3 and pi over 3,
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tan x maybe looks
something like this.
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Yeah, you know, give or take.
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OK, so what's
important about this?
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Well tan x is an odd function.
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And this interval from-- this
is minus pi over 3 to pi over 3,
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and this is the
curve y equals tan x.
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So this is an odd function.
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So the integral in question
is this positive area
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minus this negative area.
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But since it's an odd
function, it's symmetric.
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So those two cancel
out perfectly.
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So for the second one, you
could do the same process
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that we did.
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But you can also use a little
bit of geometric reasoning
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to realize that this second
one is just equal to 0.
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In either case, doing
either of these integral
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with Riemann sums would
be really atrocious.
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So, in both cases, what
we're really happy about
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is that we have these
other tools that
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allow us to evaluate definite
integrals without going back
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to the Riemann definition
of the integral.
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So, I'm going to stop there.