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PROFESSOR: Hi.

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Welcome back to recitation.

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In lecture you've been
learning about applications

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of integration.

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And one of them
is how to compute

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average values of a function.

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So I have a nice average
value problem for you here.

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So Christine was out jogging
one day and she saw a bear,

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and so she broke into a sprint.

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She starts speeding up,
and her velocity t seconds

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after she started sprinting
is given by v of t

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equals 1500 divided by
the quantity 100 plus

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t minus 5 squared, the
whole thing minus 7.

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And it's in meters per second.

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So OK, so she starts
at t equals 0,

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she's going 1500 divided by
125, so that's 12 minus 7,

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so 5 meters per second.

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And she starts to speed
up and after 5 seconds

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she's going 1500
divided-- well this is 0--

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so she's going 8 meters
per second at 5 seconds.

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And then she looks
over her shoulder

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and she sees, oh no,
it wasn't up a bear,

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it was just an unkempt
math graduate student,

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so then she starts
to slow down again.

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She slows back down and
10 seconds later she's

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back to her original velocity.

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So the question is, over the
10 seconds of this sprint,

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she sped up, she slowed
down, from 5 meters

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per second to 8 meters per
second and then back to 5.

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What was her average
velocity over this time?

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So why don't you pause
the video, take some time

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to work that out, come back
and we can work on it together.

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So hopefully you had some
luck working on this problem.

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I have over on my left here
a little picture of the graph

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of this function v of t.

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I described it to you
earlier, but just you

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see, OK, so it starts at 5,
after 5 seconds it's at 8,

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and then it goes back
down to 5 at 10 seconds.

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So this is roughly
the picture, so we

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have a feel for what
we're working with.

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Now what we want to compute
is the average velocity

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over this time.

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So anytime you have to
compute an average value

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of any function
what you want to do,

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you always do the same thing.

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So you want to compute the total
contribution of that function

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over the interval in question.

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So you want to add it up,
you want to integrate it

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over the interval in question.

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But then you want to divide
by the length of the integral.

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So in particular, when you
have a velocity function what

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that means is, you
integrate the velocity.

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So that gives you the
total distance traveled,

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and then you divide
by the time interval.

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S you're just taking total
distance divided by total time.

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So in this case we can write
down this average value.

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So I'm going to use
just a made up notation.

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I'm going to write avg, for
average, of v, so what is it?

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Well the first
thing I have to do

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is I want to not
forget to divide

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by the length of
the interval here.

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So in this case, the
interval is from 0 to 10,

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so it has length 10
minus 0, which is 10.

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So I want 1/10 in front and
what I want to multiply this by,

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is I want to multiply
by the integral

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over this entire interval
of my function v.

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So I just take the formula that
I had over there for v and I

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plop it down into my integral.

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So it's the integral
from 0 to 10

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of 1500 divided by 100 plus
t minus 5 quantity squared

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minus 7 dt.

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So this integral is the average
value that we're looking for.

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OK.

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So now we have to go
about evaluating it.

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Now, this looks kind of ugly,
but actually it's not that bad.

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I mean, one thing you can
notice, so the minus 7 part,

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that's going to be
easy to take care of.

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And then in this more
complicated part, well

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there's a kind of obvious
substitution here.

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You can set u equal
t minus 5, that'll

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simplify it a little bit.

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and then what you'll
see is that you

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have an integrand of the
form something divided

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by 100 plus u squared.

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And so that's 10
squared plus u squared.

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So that should keep-- and
the something as a constant.

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So that should put you in the
mind of a tangent substitute,

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or just even not
by substitution,

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you can just
remember that that's

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an arctangent type of thing
that you're going to get out.

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so I'm not going to go through
all the steps of doing that,

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but we should get at the end--
well OK, so I've got this 1/10

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out in front, so I have
to not forget about that.

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I get 1/10 times,
I've gone ahead

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and I've computed
the anti-derivative

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and it's 150 times arctan of
the quantity t minus 5 over 10.

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So here's that t minus 5, that's
the same t minus 5 over there

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that's coming out.

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Minus 7t.

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And you want to take that
between t equals 0 and t equals

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10.

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So this is, I've
just, I've gone ahead,

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I've computed the
anti-derivative for us.

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There's my 1/10 because
I'm doing an average value

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and then I'm, it's
a definite integral

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and so I'm using fundamental
theorem of calculus;

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there are my bounds.

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OK, so now we just have
to plug in and evaluate.

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So this is equal
to, so let's see.

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So we've got when we put in
t equals 10-- well the 1/10

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and the 150 gives me a 15--
arctan of 1/2 minus-- well it's

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1/10 of 7 times 10, it's just 7.

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That's from the first one.

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And from the second
one I guess minus--

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so it's going to be 15
arctan of minus 1/2.

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When I put in t equals 0,
and when I put in t equals 0,

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the 7t is just 0.

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OK.

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This is a little messy.

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We can simplify it a little bit,
because, remember arctangent is

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an odd function.

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So arctan of minus x is
minus arctangent of x.

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So minus 15 arctan minus 1/2 is
the same as just 15 arctan 1/2.

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And so we can combine those.

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So we can rewrite that.

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Let's go all the way over here.

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We can rewrite this as
30 arctan 1/2 minus 7.

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So that's about it as nice a
form as you can make this take.

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If you wanted to plug
it into a calculator,

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I think you'd see that this
is approximately equal to 6.9

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and I guess the units there are
going to be meters per second.

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So her average
velocity over this time

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is about 6.9 meters per second.

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Let's just quickly
recap what we did.

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So we started all
the way back here

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with this velocity
function v, and we

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wanted to compute it's average
value over the interval 0

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less than or equal to t,
less than or equal to 10.

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So we did what we always do
in problems of that sort.

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So we came over
here, and whenever

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you want to compute
an average value,

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you take 1 divided by the length
of the interval in question,

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times-- then just you
integrate the function who's

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average value you want
over the interval.

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OK?

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And so then we computed, well
we didn't show any of the steps,

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but we computed the
definite integral

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by taking the anti-derivative
and plugging in

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and it happened that the answer
worked out to be about 6.9.

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So there you go.

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I'll stop there.