A Magical Theorem, Part II: Areas by way of Antiderivatives

So how are areas related to antiderivatives?  In other words (and symbols):

fundamentalquestion

This is the fundamental question.  When you know the answer, you will know the “Fundamental Theorem of Integral Calculus”.

In your first physics class, while studying kinematics, you learned that the area under a velocity vs time graph tells us the change in the object’s position.  To help jog your memory, play the video below:

The upper graph shows the velocity as a function of time for an object that happened to have a constant velocity of 10 m/s.

The lower graph shows the position vs time for the same object.  The velocity is constant so the position graph is linear with a constant slope.

On the velocity graph, you can see the area being swept out from t = 3 s to t = 7 s.  And on the position graph, you can see how far the object has moved during the same time interval.  Try stopping the animation at different times. Examine the graphs until you have convinced yourself of the following claim:

For a given time interval, the area under the Velocity vs. time graph will be equal to the change in position during that time interval.

 

YOU MAY BE SKEPTICAL…

After all, that was just with constant velocity.  But what if the velocity is changing?  Like so…

The velocity is increasing linearly and the position is increasing quadratically. [Remember, velocity is the derivative of position with respect to time. And the derivative of a 2nd-degree polynomial is a linear function.]

But even so, once again we see:

For a given time interval, the area under the Velocity vs. time graph will be equal to the change in position during that time interval.

STILL SKEPTICAL? STAY TUNED…

The velocity does not have to be changing linearly.  In fact, it does not even have to be velocity and position that we are talking about.  It can be any two quantities where one is the derivative of the other. Here’s one where the first graph is the rate at which water flows into a container and the second graph is the volume of water in that container:

Stop the animation at various points to check our claim:

If f(x) is the derivative of F(x), then the area under the graph of f(x) is equal to the change in F(x) over that same interval.

 And that answers the question we started with. What does area have to do with antiderivatives?  The area under the graph is equal to the change in the antiderivative.  That is called the Fundamental Theorem.  We write:

fundamentalanswer

WHY IS THIS USEFUL?

Well, suppose you want to find the area under some graph. This theorem gives you a procedure you can follow:

Step 1: Find an antiderivative.

Step 2. See how much the antiderivative changed by over the interval.

For example, suppose we want to know the area under the graph of f(x)=x2 between x=1 and x = 2, as illustrated below.

defintexample

Let’s see.  Can we find an antiderivative?  Oh, yes, this one we did already.

F(x)=(1/3)x3 is an antiderivative of  f(x)=x2.

So now we just have to see how much F(x) has changed over the interval from x = 1 to x = 3.

F(3) = (1/3)(2)3 = 8/3

F(1) = (1/3)(1)3 = 1/3

F(3) – F(1) = 7/3

But what if I can’t find the antiderivative?

Well, that could be a problem.  That’s why you will spend so much time developing this skill in AP Calculus.  But I will reassure you again: in AP Physics, the antiderivatives you need will be manageable.  And there’s always Wolfram Alpha

I still don’t see why this is so important?  Are there lots of areas to be found in AP Physics?

Hmm.. No and yes.  See next post…

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