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 for the

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last couple of lectures.

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And you've just started
with the

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fundamental theorem of calculus.

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So, that gives us this whole
powerful tool to compute a

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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

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over 3 of tan x dx.

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So why don't you pause the
video, take a couple minutes

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to work on those, come
back and we'll

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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 of

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definite integrals here.

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So one tool for definite
integrals is Riemann sums. So,

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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 and

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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 is using this

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great tool that we have,
which is the

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fundamental theorem of calculus.

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So the fundamental theorem of
calculus says, when you have a

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definite integral, instead of
computing Riemann sums, what

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you can do is compute an
antiderivative and use the

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antiderivative to find out
what the integral is.

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Just compare two values with the
antiderivative and that'll

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give you the value of the
definite integral.

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So, in particular, let's do this
first integral first. So

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we have the integral from
pi over 6 to pi

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over 3 of tan 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 says that the value

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of this definite integral, in
order to compute it, we just

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take 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 this is equal to

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ln of the absolute value of
cosine x for x between pi over

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6 and pi over 3.

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So, OK, so this, you know, this
is this notation that

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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

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cosine of pi over 3 minus ln
of the absolute value of

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cosine of pi over 6.

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Now, we could leave our answer
like this but we can also

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manipulate it and put it
in a little nicer form.

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So, cosine of pi over 3 is--

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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--

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OK, so 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

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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

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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 to combine

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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 for the

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same number, which
is the value of

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this definite integral.

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OK, so, for the second integral
now, we can do

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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, you'll

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take the difference in values.

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We can also do something
a little bit

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clever for this one.

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Which is, definite integrals
have some geometric

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interpretation, right?

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And if you think about
the function tan x--

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so we could put up a
little graph here--

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so in between minus pi over 3
and pi over 3, tan x maybe

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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--

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this is minus pi over 3 to pi
over 3, and this is the curve

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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 minus this

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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

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process that we did.

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But you can also use a little
bit of geometric reasoning to

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realize that this second
one is just equal to 0.

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In either case, doing either of
these integral with Riemann

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sums would be really
atrocious.

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So, in both cases, what we're
really happy about is that we

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have these other tools that
allow us to evaluate definite

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integrals without going back to
the Riemann definition of

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the integral.

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So, I'm going to stop there.