Interpreting definite integral of rate function

- [Instructor] Julia's
revenue is r of t thousand

dollars per month, where t
is the month of the year.

Julia had made $3,000 in
the first month of the year,

what does three plus
the definite role from

one to five of r of t dt equals 19 mean?

And we have some choices.

So like always, pause the video and see

if you can work through it.

Alright, now let's work
through this together.

So they tell us that she made
$3,000 in the first month

and we also see this three
here, so that's interesting.

Maybe they represent the same thing,

we don't know for sure yet.

But let's look at this definite integral.

The definite integral from one to five

of r of t dt, this is the
area under this rate curve,

r of t is the rate atwhich
Julia makes revenue

on a monthly basis.

So if you take the area
under that rate curve,

that's going to give you the net change

in revenue from month one to month five,

how much that increased.

And so if you add that to the amount

she made in month one, well that tells you

the total she makes from
essentially time zero

all the way to month five.

And they're saying that is equal to 19.

So let's see which of these choices

are consistent with that.

Julia made an additional $19,000 between

months one and five.

Choice A would be correct if you didn't

see this three over here.

Because just the definite
integral is the additional

between months one and
five, but that's not

what this expression says,
it says three plus this

is equal to 19.

If it said Julia made
an additional $16,000,

well that would make
sense because you could

subtract three from both sides
and you'd get that result,

but that's not what they're saying.

Julia made an average
of $19,000 per month.

Well once again, that's also not right

because we just said from the beginning,

from time zero, all the
way until the fifth month,

she made a total of $19,000.

Not the average per month is $19,000.

Julia made $19,000 in the fifth month.

Once again, this is not just saying what

happened in the fifth month.

This is saying, we have the
$3,000 from the first month

and then we have the additional from month

between month one and month
five, so that's not that.

So this better be our choice.

By the end of the fifth
month, Julia had made

a total of $19,000.

Yes that is correct.

She made $3,000 in month
one and then as we go

between month one to
the end of month five,

to the end of the fifth
month, she has made

a total of $19,000.

Let's do another one of these.

So here we're told the function k of t

gives the amount of ketchup in kilograms

produced in a sauce
factory by time in hours

on a given day.

So this is really quantity
is a function of time,

it isn't rate, what does
the definite integral

from zero to four of k
prime of t dt represent?

Once again, pause the video and see if you

can work through it.

Well k of t is the amount of ketchup

as function of time.

So k prime of t, that's going to be the

rate at which our amount of ketchup

is changing a function of time.

But once again, when
you're taking the area

under the rate curve, that
tells you the net change

in the original quantity
in the amount of ketchup.

And as the net change between
time zero and time four,

so let's see which of these
choices match up to that.

The average rate of change of the ketchup

production over the first four hours.

No, that does not tell us
the average rate of change,

there's other ways to calculate that

the time it takes to
produce four kilograms

of ketchup, so does this
represent the time it takes?

To represent four kilograms of ketchup.

No, this four is a time right over here.

This is gonna tell you how much ketchup

gets produced from time zero to time four.

The instantaneous rate of
production at t equals four.

Now this would be k prime of four,

that's not what this integral represents.

The amount of ketchup produced over the

first four hours, yep
that is exactly right.