Presentation on theme: "Getting the Most Out of the Fundamental Theorem Dan Kennedy Baylor School Chattanooga, TN"— Presentation transcript:
Getting the Most Out of the Fundamental Theorem Dan Kennedy Baylor School Chattanooga, TN firstname.lastname@example.org
One of the hardest calculus topics to teach in the old days was Riemann sums. They were hard to draw, hard to compute, and (many felt) totally unnecessary.
That was why most of us quickly moved on to antiderivatives, which is how we wanted students to do integrals. Needless to say, when we came to the Fundamental Theorem, students found it to be the greatest anticlimax in the course. Integration and differentiation are reverse operations? Well, duh.
Then along came the TI graphing calculators. Using the integral utility in the CALC menu, students could actually see an integral accumulating value from left to right along the x-axis, just as a limit of Riemann sums would do:
So now we can do all kinds of summing problems before we even mention an antiderivative. Historically, that’s what scientists had to do before calculus. Here’s why it mattered to them:
The calculus pioneers knew that the area would still yield distance, but what was the connection to tangent lines? And was there an easy way to find these irregularly-shaped areas?
Since the time of Archimedes, scientists had been finding areas of irregularly-shaped regions by dividing them into regularly-shaped regions. That is what Riemann sums are all about. 2.0332812.0082482.000329 With graphing calculators, students can find these sums without the tedium. They can also imagine the tedium of doing these sums by hand!
Best of all, they can actually see the limiting case: And the calculator shows the thin rectangles accumulating from left to right – ideal for understanding the FTC!
Let us consider a positive continuous function f defined on [a, b]. Choose an arbitrary x in [a, b]. x
Each choice of x determines a unique area from a to x, denoted as usual by
Thus is a function of x on [a, b]. What is the derivative of this function?
This was the FTC. This was the result that changed the world. 2.000329 Now, instead of wasting a full afternoon just to get an approximation of the area under one arch of the sine curve, you could find one antiderivative, plug in two numbers, and subtract!
Since 2000, the AP Calculus Test Development Committee has been emphasizing a conceptual understanding of the definite integral, resulting in these “new” problem types: Functions defined as integrals Accumulation Problems Integrals from Tables Finding, given and Interpreting the Definite Integral
This problem had been checked: 1.by the author who had written it; 2.by the committee that had okayed it; 3.by the committee that had okayed it for a pre-test; 4.by the ETS test development specialists; 5.by our committee, reviewing the final form of the college pre-test. My colleagues were two problems further into the test when I asked if we could go back for another look.
The proposed key was (B). That is, While everyone was concentrating on the Fundamental Theorem application, they had missed the hidden “initial condition” that y must equal zero when x = 1!
(c) Since is positive from -6 to -1 and negative from -1 to 4, the minimum occurs at an endpoint. By comparing areas, h(4) < h(-6) = 0, so the minimum occurs at x = 4. This “area comparison” genre of problem was pretty common in the early graphing calculator days.
Accumulation Problems Perhaps the most groundbreaking change in the 1998 AP Course Description was the decision not to list the “applications of integration” that a student should know. Instead, students would be expected to have enough familiarity with “accumulation” problems to model them with integrals in fresh situations. The exams since then have provided an abundance of fresh situations!
All students should know how to interpret the following applications as accumulations: Areas (sums of rectangles) Volumes (sums of regular-shaped slices) Displacements (sums of v(t)∙∆t) Average values (Integrals/intervals) BC: Arclengths (sums of hypotenuses) BC: Polar areas (sums of sectors)
Problem of the Day #27: I give this before the FTC and before any definition of average value.
Answer: Take the area of the circle and divide it by the diameter! Average chord length =.
Another implication of the Fundamental Theorem (and a source of several recent problems that have caused trouble for students): Thus, given f(a) and the rate of change of f on [a, b], you can find f(b).