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## Integral Calculus

### Course: Integral Calculus > Unit 1

Lesson 1: Accumulations of change introduction# Worked example: accumulation of change

AP.CALC:

CHA‑4 (EU)

, CHA‑4.A (LO)

, CHA‑4.A.1 (EK)

, CHA‑4.A.2 (EK)

, CHA‑4.A.3 (EK)

, CHA‑4.A.4 (EK)

An example relating rates of change with a leaky bathtub. Created by Sal Khan.

## Want to join the conversation?

- What if the graph was a curve? why is the integral of the function equal to the area function of the curve? like why is the area function of y=2x be x^2? i don't get how integrating a function can give you the area of the function :((14 votes)
- Really Great Questions. It is just as great that you are thinking this way! Your questions will be answered in the very next section of videos called "Riemann Sums", which will guide your intuition of how when dx, which can be thought of as length along the x axis, times f(x), which can be thought of as the height of the graph above the x axis at x, are multiplied together, they form a width x height = area argument the represents the area of the region under the curve. There is more to it than that, but the Riemann Sum videos will explain it in more detail.

Keep Studying!(19 votes)

- How can you find the area under R(t) when it is discontinuous?(18 votes)
- my intuitive response would be to calculate within bounds.(from -to) and then add up.(4 votes)

- Why did you multiply all of those numbers by 1/2? And why did you use 40 minutes?(3 votes)
- He didn't multiply all of the numbers by 1/2--only the ones which represented the area of a triangle! Remember, the area of a triangle is 1/2 * base * height. He used 40 minutes in his last two calculations because that was the width given.(8 votes)

- The question doesn't indicate what happened to the leak after 20 minutes, by the time the drain valve was opened, the leak was contributing 2 gals / minute as a minimum, but the rate of change on that leak had been linearly increasing at a rate of 1 gal/minute for each 10 minutes leading up to this point.

As the plot does not continue beyond 20 minutes, I calculated both options, with the actual amount of water being in between the two points.

How did the leak stop after the drain valve opened, but before the tub was empty, and how are we to know that this should not be projected going forwards?(4 votes) - What would the area under the curve represent if the Y axis was NOT a rate?

ie. a temperature vs time graph or a position vs time graph.(1 vote)- The integer always represents the accumulation of the quantity represented by the y-axis over the quantity represented by the x-axis. There are some that are very easy to understand, and other that apparently makes no sense.

In your examples, if the x-axis is time and the y-axis is temperature, the integral would represent the accumulation of temperature that happen in a certain time, this could be useful if you are dealing with a material that absorbs heat and want to know how much heat it has absorbed over a period of time.

Your second example is more complex to find a useful representation, if the x-axis represents time and the y-axis represents position, then the integral represents the total accumulated "position" that something accumulated over time. I really can't think of a useful interpretation of that one.(5 votes)

- i tried solving problem myself using Riemann sums (left rule) using rectangles of width 5 for both the curves and i got the answer to be close to 600 gallons.. i realized that the difference should be 90 gallons because the value of the function at t=20 is 2 and not 20. where am i going wrong?(2 votes)
- First off, remember that using a Riemann sum with intervals of width 5 is going to lead to an
**approximation**which may be fairly far off from the actual area under the curve. If I write a left-hand approximation using 12 rectangles of width 5, here's what I get:

L_12 = 5(0 + 0.5 + 1 + 1.5 + 2 + 18.75 + 17.5 + 16.25 + 15 + 13.75 + 12.5 + 11.25) = 550

So it does underestimate the exact value, in large part because you are leaving off that highest point. Does that help?(6 votes)

- How would you model the level of water in the bathtub if water was also being put in at different rates?(3 votes)
- At4:30, Sal says that the area of the trapezium or trapezoid is average height times the base, but from I recall that the area of a trapezium or trapezoid is (upper bound+lower bound)*height*0.5? Where does the area formula for the trapezium Sal had stated come from?(2 votes)
- Sal chose to divide the trapezoid into a square and a triangle - both easy to compute the area in your head.

He could have used your formula, but then we would have to imagine the trapezoid laying on its side since what you call the upper and lower bound need to be parallel to each other. In that case the upper bound is 20, the lower bound is 10 and the height is 40 giving

(20+10)(40)(1/2) = (30)(40)(1/2)=(30)(20)=600, the same result.(2 votes)

- The slope of the line is positive between t=0 and t=20. Meaning the rate of gal/min is positive. My understanding is that the tank is being filled over the first 20 minutes.

Then, between t=20 to t=60 the slope is negative. My understanding is that this is when the draining of water begins. So, to answer the problem, I would calculate the area under the curve between t=20 and t=60.

In the video (4:39), to find the total water drained the areas under both curves were added.

My question: please explain where is the flaw in my understanding of the problem.(2 votes) - Is it correct to say that the function here, r(t), is really just the first derivative of the anti-derivative we are looking for here? Could we find an anti-derivative of r(t) algebraically, call it R(t), and then evaluate R(0) to answer this question?

If this is correct, how do we do this?(1 vote)- That's actually the logic behind a definite integral, which is explored later in this unit, except that we need to measure the "accumulation of change" of r(t) from 0 to 60, which isn't R(0). In fact, since any function has infinitely many derivatives, R(0) doesn't actually have a definite value. To do what you're describing, we need to define r(t) as a piecewise function:

r(t) = .1t from [0,20] and -.25t+25 from (20,60]

The area under r(t), or net change, is then the definite integral from 0 to 20 of .1t dt + the definite integral from 20 to 60 of -.25t+25 dt. This value can be evaluated with the antiderivatives of both parts of the piecewise function. This value is actually equivalent to the area Sal finds geometrically in this video.(3 votes)

## Video transcript

Voiceover: It took 20
minutes before Jughead noticed that his hot
tub had sprung a leak. Once he realized it, he
opened the hot tub's drain and the rest of the water
rushed out in 40 minutes. The rate at which the water
drained from the hot tub in gallons per minute is shown. How many gallons of
water were in the hot tub before it started to leak? So let's see what they have over here. So they've plotted gallons per minute versus time in minutes. So we see here, this blue line, this is the rate at which
water drained from the hot tub. So at minute zero, the
water wasn't really draining from the hot tub and then
not just more drained but the rate at which the
water was draining increased. So 10 minutes into his
bath, the water was draining at a gallon per minute and
then 20 minutes into his bath water was draining at
two gallons per minute. Then he noticed it and he opens the drain. I guess he wants to accelerate
the end of his bath. So he opens the drain
and then all of a sudden water starts draining out
at a much higher rate, at 20 gallons per minute,
but then that decreases. So we could think about physically
why that might decrease. Maybe there was just less
pressure or whatever. We're not going to go
into the physics of it, but we're just going to
take this chart as fact. The rate at which the
water drains decreases all the way to the 60th minute, which is 40 minutes after
he opened the drain. At the 60th minute, all of the water was actually drained out. So given that, how do we think about how many gallons of
water were in the hot tub before it started to leak? And I encourage you to pause the video and try to see if you could
figure it out yourself before I work through it. Well, to answer this question: How many gallons of
water were in the hot tub before it started to leak? That is answering the same question as How much total hot water drained out? So how much drained out? And to think about that,
we can just kind of go back to what we knew before we
learned about calculus. If I have something
happening at a fixed rate, so let's say that this
is gallons per minute. So this is still the same
context, I guess you could say. This is time in minutes. Let's say things are draining
out at a constant rate. If you wanted to figure
out how much drains out over a certain interval of time, let's say this interval
of time right over here, let's call that delta t, you would just multiply the
rate over that interval of time, which we could represent by this orange height right over here, times the amount of time that passed by, which would give you the area under the curve over that interval. the area under the curve
over that interval. This area would tell you the gallons that drained over that delta t. And this doesn't just apply
when you have a constant rate. If your rate looked something like this, as we've seen in other videos, you could figure out the
amount that has drained in a specific interval, let's say this interval right over here, by essentially figuring out
the area under the curve over that interval,
and the units work out. If you multiply gallons
per minute times minute, this area is going to
be in terms of gallons. Another way you could think about it, and this goes back to,
well, how do you figure out the area of a trapezoid right over here. Well, the area of a trapezoid, you could find the average
height of the trapezoid, which would be the average of the beginning and the end period, you take that average over there, and this would work for a line like this, if you take that average height and multiply by the change in time, you are going to figure out that area. And that is another way
of thinking about it. You are taking the average
rate over that interval times the interval, and
that is going to give you the total number of gallons. And so we just have to
apply that idea over here. We just have to literally figure out the area under the curve
over the entire interval when the water was actually
leaking or draining. So essentially the area under the curve between zero minutes and 60 minutes. And so it is going to be this area plus all of this area under this part as well. To help us think about that, I'm going to just split
that up into some sections. So I'll have this
triangular section up here. I could just think about
this as a trapezoid, but I'm just going to split it up into a triangle and a rectangle. And then I have this section
right over here in green. So what is the area of this entire thing? Well, the area right over here, we have 20 minutes times two
gallons per minute times 1/2. This gives us the area
under this triangle. So that's going to be 20 gallons. We see that the units work out nicely. So that is essentially
how much has drained out in the first 20 minutes. And then this green area
is going to be 40 minutes times 10 gallons per minute. And actually maybe you know the units, since I'm breaking it
up in this strange way, I'll just figure out this
area in a unit-less way. So 40 times 10, which is equal to 400. And then, finally, in blue,
I have 40 times this height right over here between
10 and 20 is another 10 but then I'm going to
multiply that times 1/2. So it's going to be 40 times 10 times 1/2, which is going to be 200. And so when you add all
of these areas together, you are going to get 400
plus 200 is 600 plus 20. You are going to have 620 gallons. 620 gallons is how much
water in total drained out or how much water was in the
tub before it started to leak, or the hot tub, yeah this is
quite big for a regular tub, bu this could be like a Jacuzzi
or a hot tub or something.