If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Area under rate function gives the net change

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.D (LO)
,
CHA‑4.D.1 (EK)
,
CHA‑4.D.2 (EK)
,
CHA‑4.E (LO)
,
CHA‑4.E.1 (EK)
If you have a function representing rate, what does the area under its curve represent?

Want to join the conversation?

  • mr pants teal style avatar for user Cameron Swartzell
    At Sal giggles as his example assumes an Instantaneous acceleration to double something's speed; Light travels at nearly half speed through a diamond (ok, a little less than), and of course at c through a vacuum, with presumably an instaneous acceleration. Sal's example makes sense if it is measuring a beam of light which travels through a diamond for the first two seconds, then a vacuum from then on.

    This diamond would have to be 299 792.458 kilometers thick.
    (48 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Bertrand Dechant
      well even if the light analogy breaks down with given math, we couldn't experimentally determine the rate at which light accelerates between the surface of the diamond and vacuum. This is due to nature but our inadequacy is modeled by Heisenberg's uncertainty principal. basically it's too low mass and moving way too fast to know velocity and position at the same time. TLDR: Only mathematically impossible.
      (17 votes)
  • aqualine seed style avatar for user MaverickFernandez
    This may be a dumb question, but at , when he is talking about something's velocity jumping from one value to another instantaneously being unrealistic, wouldn't one thing hitting something else, for example a baseball bat hitting the ball cause the ball's velocity to jump instantly?
    (7 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user naveensjoseph14
      hi,
      in case of instantaneous velocity you can't get the exact instant. its just a convention. we find it using avg velocity between extremely small time interval. lets say an intreval between 5.00000000001 and 4.99999999999999 can be used in such cases. that is why we use limits there.
      (4 votes)
  • leaf red style avatar for user Ngoc Nguyen
    How do you know or decide that time will be the x axis and rate will be the y axis? Can it be the other way around, rate on x axis and time on y axis ? What if I want to make distance (d) x axis or y axis? Thanks
    (4 votes)
    Default Khan Academy avatar avatar for user
    • leafers tree style avatar for user Guess Again
      Your x-axis will typically correspond to the independent variable. In situations such as those shown above, this will often be time, as time is continuous and does not depend on anything. Therefore, the dependent variable will typically be on the y-axis. In the cases shown here, rate is dependent on time (just have a look at the units), so it will go on the y-axis.

      In terms of distance, that is something that can change over time (assuming time is your other variable), so it should go in the y-axis.
      (10 votes)
  • piceratops sapling style avatar for user Krutik Desai
    The rate here means speed, right?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • leaf grey style avatar for user Nabla ∇
      The term rate is used instead of speed (or velocity), because is a more general one.

      It can be a change of position over a change in time, but it can also be any change that occurs over an interval of time (height, heartbeats, shoes sold, etc.).
      (7 votes)
  • old spice man green style avatar for user Donepudi Aditya
    What is RATE in calculus? I really confuse with this term.I am not able to understand it.Please help by giving an accurate definition with example.
    (3 votes)
    Default Khan Academy avatar avatar for user
    • starky ultimate style avatar for user Epiksalad
      Rate doesn't mean anything specific or special in calculus. As usual, it means what rate normally means - how much of an activity is done per unit time.

      For example, if a body is moving at 5 meters per second (as told by Sal at ), it moves a distance of 5 meters over a period of 5 seconds.

      If you want a more 'accurate' definition,
      In mathematics, a rate is the ratio between two related quantities. If the denominator of the ratio is expressed as a single unit of one of these quantities, and if it is assumed that this quantity can be changed systematically (i.e., is an independent variable), then the numerator of the ratio expresses the corresponding rate of change in the other (dependent) variable.
      The most common type of rate is "per unit of time", such as speed, heart rate and flux.

      I hope you are satisfied.
      (6 votes)
  • spunky sam blue style avatar for user Muhammed Hafez
    why didn't we multiyply 5*2 directly in the second problem ?? .. thanks in advance
    (2 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user doctorfoxphd
      Well, we cannot do that because the velocity is not a constant 2 over the 5 seconds of elapsed time. It is a "piece-wise function" that has one rate over the first two seconds and one rate over the last three seconds. So, we have to calculate it in two parts:

      Two seconds were spent at 1 m/s and the final three seconds were spent at 2 m/s.
      2s*1 m/s + 3s*2 m/s = 2 + 6 = 8 meters

      Notice that 2*5 would give the wrong answer.
      (6 votes)
  • female robot ada style avatar for user Raviv Sarch
    what even is the advantage of finding the area in this situation?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Stefen
      There is no advantage other than being a demonstration that rates and rates of change (differential calculus) can be summed (integral calculus). The tutorial foreshadows the connection between rates and areas under curves. You will soon see that this connection is the foundation of the Fundamental Theorem of Calculus.
      (6 votes)
  • blobby green style avatar for user bharshbarger
    The integral of a function gives you the area under the function above the x axis. Ok. But why? Say I have a parabola Y=x squared and want to find the area under it from 0 to 1...what is about X cubed divided by three that enables this calculation? I teach this, I don't like telling students that, it's area under the curve...there's nothing logically appealing about x cubed divided by 3. Now multiplying the rate of change by x, this is okay...but that's not what you are doing in an integral where the function is now divided by 3, in my example. Got a video on this?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf grey style avatar for user Qeeko
      You should have a look at an elementary development of the integral to see why this is the case. The following book by Apostol has a clear exposition that you might find interesting. Do note, however, that this PDF version of the book contains numerous misprints and poor mathematical typesetting (this is not so if you buy an actual copy of it), so beware and read critically. Any notation you (possibly) don't understand is probably covered in the introduction chapter.

      The pages 2 (from "Historical background") to 10 is an informal (but motivating) treatment of your example, area of a parabola segment.

      Chapter one, "The concepts of integral calculus", is also worth a read. If you are only interested in the problem concerning area, read till you get to Theorem 1.10 on page 75, which proves that the integral indeed gives the desired area (in the proof of this theorem, the letter Z that appears should actually be an I - a misprint).

      The book (PDF):
      http://www.mif.vu.lt/~stepanauskas/AM1/Tom%20Apostol%20-%20Calculus%20vol.1%20-%20One-variable%20Calculus,%20with%20an%20Introduction%20to%20Linear%20Algebra%20(1975).pdf
      (3 votes)
  • aqualine seedling style avatar for user Erik Hillberg
    Were did he get 5s from in the 2 problem?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • starky sapling style avatar for user 20leunge
    Is finding the definite integral between two points for a velocity function the same thing as finding the secant line that goes through those two points in the corresponding position function?
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Voiceover] Let's say that something is traveling at a constant rate of five meters per second. That's its velocity in one dimension. If it was negative, we'd be moving to the left. If it's positive, it's moving to the right. Let's say that we care about what is our change in distance over, the delta symbol represents change, over a change in time of four seconds. Over four seconds. I could say from t equal zero to t is equal to four. That's our change in time. That's our four-second interval that we care about. Well, one way to think about it is what a rate by definition is nothing but a change in some quantity. In this case, it's distance over a change in some other quantity. In this case, we're thinking about time. Or another way to think about it, if we multiply both sides times change in time, you get your change in distance is equal to your rate times change in time. This is very close to you, you might remember from pre-algebra, distance is equal to rate times time. Time, that just comes from the definition of what a rate is. It's a change in one quantity with respect to another quantity. If you just apply this, you can say, "Okay, my rate is a constant five meters "per second and my delta t is four seconds." So times four seconds. Well, that's just going to give you 20. That's just going to get you 20. Let me do that in same color that I had for the change in distance. That's going to be 20. Then the seconds, cancel the seconds, 20 meters. So my total change in distance over those four seconds is going to be 20 meters. Nothing new here. Nothing too fancy. But I want to do now is connect this to the area under the rate function over this time period. So let's graph that. Let's graph it. That's my rate axis. This is my time axis. This is going to be in seconds. This is going to be in meters per second. Let's see, one, two, three, four, five. Let's see if it's about enough, and then I go one, two, three, four, five. Our rate, at least in this example, is a constant, is a constant five meters per second. It's a constant five meters per second. That is my r of t in this example. What did we just do here? We just multiplied our change in time times our constant rate. We just multiplied our change in time. So from time equal zero seconds to four seconds. It's this length here, if we think on that axis. Now we multiply it times our constant rate. We multiply it times this right over here. If I multiply this base times this height, what am I going to get? I'm going to get this area under the rate function. Now that area is going to be 20. If we went with the units of them, obviously you're used to things of area being something units squared because it's usually meters times meters or miles times miles or inches times inches so it'll be inches squared, meters squared, or miles squared. But here if we go at the units of the axis of the meters per second times seconds which is going to get you meters. But the important thing here is that the units here or the area here is 20. So this for that very simple example, it looks like the area under the rate curve is equal to the net change over that time period where the rate is something with respect to time. Well let's test the little burrow. It's just going to be more intuition here. Let's say that we have a different rate function. Let's say that we will make with the different, let's say instead we had a rate function. I'll use on this yellow. Let's say on a rate function, that is... Let me make it a little bit interesting. Let's say it's one meter per second, for our time is zero is less than or equal to time which is less than or equal to two seconds. Honestly, this is all in seconds where we're of time. There's two meters per second, for t is greater than two seconds. What's that going to look like? Actually, try to graph it yourself, and just say, "Well, what is the total change in distance "over the first, let's say five seconds?" We want to do delta t over, not the first four seconds but the first five seconds. Well, let's graph it. Let's graph it. So this is one meter per second. One meter per second. That is two meters per second. That's in meters per second. That's my rate axis. This right over there is going to be my time axis. One, they're obviously not of the same scale, three, four, five. What is this rate function look like? Well, my rate is one meter per second between time is zero and two, including two seconds, and then the rate jumps. Nothing can accelerate instantly like this. You'd need an infinite force or an infinitely small mass I guess to, or maybe there's something that's thinking about... Anyway, I only get two complex here but this is unrealistic mode. It's not typical for something to just have spontaneous velocity increase like that but let's just go with it. Then after the two seconds, we are at a constant rate of two meters per second. Now, what is our total change in distance over the first five seconds here? So here, we care about the first five seconds, or we can break up the problem. We could say, "Well, over the first two seconds." Change in time is two seconds, times our constant rate over those two seconds. It's going to be two seconds times one meter per second. Well, that's going to give us two meters. So this here is going to be, shall we do that in orange color, that's going to give us two meters there, and then we look at the next section. Our change in time here is three seconds, and then we multiply that, times our constant two meters per second. That's going to give us an area of six. If we look at the units, in both cases, we're multiplying seconds times meters per second which is going to give us meters. This is going to be two plus six meters or eight meters. So hopefully this is giving you the intuition that the area under the rate curve or the rate function is going to give you our total net change in whatever that rate thing was finding the rate of. In this case, it is distance per unit time. If you take the area under the rate function, that kind of distance for the speed or this velocity function over some period of time, that area is going to be our total net change in distance.