SUNY Geneseo Department of Mathematics

The Fundamental Theorem of Calculus, Part 2

Friday, November 20

Math 221 02
Fall 2020
Prof. Doug Baldwin

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I don’t plan to hold weekly individual meetings during break, i.e., Wednesday through Friday next week. Some of you have recurring meetings scheduled for these days. If you want to meet with me next week, see if you can move it earlier; if you don’t see a need to meet next week, feel free to just wait until the week after break.

SI

The next session is Sunday, which will be for open-ended questions, discussion, etc.

There will be no session Tuesday next week.

Emiliana will send a survey re good times for post-break SI sessions. Expect it late in the break or immediately after.

Fundamental Theorem Part 2 (Evaluation Theorem)

From “The Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem” and this discussion.

The Basic Idea

Some examples with good solutions in the discussion:

Some definite integrals and their values

Some examples without solutions (yet):

The integral from 1 to 4 of √x

Start by rewriting √x as x1/2, so you can use the antiderivative power rule on it. Which is a good thing to do because the Fundamental Theorem part 2 says you can evaluate a definite integral over a certain interval by evaluating an antiderivative at the endpoints of the interval and subtracting.

Integrate square root of x from 1 to 4 by evaluating its antiderivative at 1 and 4 and subtracting

How about the integral from 0 to 1 of x + ex?

Use the same strategy from the Fundamental Theorem:

Integrate x plus e to the x from 0 to 1 by evaluating antiderivative at 0 and 1

Applications of Integration

Suppose a car accelerates in such a manner that its speed after t seconds is given by s(t) = 3t + 2 m/sec. How far does it travel in the first 5 seconds?

This question is asking about the cumulative effect of a rate of change — speed is the rate of change of distance, and the question is asking what the cumulative effect of moving at a varying speed is over 5 seconds.

Car with speed 3 t plus 1 passing one point when t equals 0 and another when t equals 5

You could think of approximately solving the problem by figuring out the speed at 0 and multiplying that by the length of some very small interval of time to get a distance traveled in the small interval, then figuring out the speed at the end of that interval, multiplying by the length of another small interval, and so on until the small intervals make a total of 5 seconds, then adding up all the distances calculated. But that process is just a Riemann sum, i.e., an integral, and we have better ways to evaluate integrals now:

Distance traveled is integral of speed from 0 to 5

Suppose water flows into an initially empty tank at a rate given by r(t) = 2sin t + 2 cubic feet per minute. How much water is in the tank after 2π minutes?

Once again, this is looking for the cumulative effect of a rate of change, i.e., it requires integrating the rate at which water flows into the tank from 0 to 2π:

Find total volume of water by integrating rate of inflow

Notice that in both of these problems the result is the amount by which something (position for the car, water for the tank) changes. It doesn’t tell you the absolute position or amount of water, unless you also know position or amount of water at the beginning (which you didn’t know in the case of the car and did know in the case of the water).

What’s the average value of f(x) = x2 over the interval [-1,1]?

This introduces the perhaps unusual idea that a continuous function can have an average value, and that the average can be calculated as an integral. One way to interpret such an average, thanks to the connection to integrals, is as the value a constant function would need to have in order to have the same area under it as the actual function does:

Average value of function between a and b is 1 over b minus a times integral of function from a to b

Using the integral formula for a function’s average value gives

Integration gives average value of x squared between minus 1 and 1 is 1 third

Next

Suddenly we seem to be able to do a lot of modeling related to cumulative results of rates of change, as long as we can find antiderivatives. Up to now the only antiderivatives we’ve been able to find are simple inverses of differentiation rules, helped out by a few laws such as the constant multiple or sum rules. But there are more powerful ways to find antiderivatives....

Please read the first, unnamed, part of section 5.5 in the textbook (from the beginning of the section up to but not including “Substitution for Definite Integrals) by class time Monday.

Please also contribute to this discussion of substitution by class time Monday.

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