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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 8

Lesson 3: Using accumulation functions and definite integrals in applied contexts

# 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?

• 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.
• 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.
• 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?
• 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.
• 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
• 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.
• The rate here means speed, right?
• 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.).
• 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.
• 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.
• why didn't we multiyply 5*2 directly in the second problem ?? .. thanks in advance
• 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.
• what even is the advantage of finding the area in this situation?
(1 vote)
• 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.
• 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?
• 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