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AP Problems Involving the Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus:. 1. If ,then . 2. 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.
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The Fundamental Theorem of Calculus: 1. If ,then 2.
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.
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. 2.033281 2.008248 2.000329 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
So But that is only half the story. Now that we know that is an antiderivative of f,we know that it differs from anyantiderivative of f by a constant. That is, if F is anyantiderivative of f,
To find C, we can plug in a: So Now plug in b:
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
(a) By the Fundamental Theorem, (b) Plug in x = 1:
This problem had been checked: • by the author who had written it; • by the committee that had okayed it; • by the committee that had okayed it for a pre-test; • by the ETS test development specialists; • The committee, reviewing the final form of the college pre-test.
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!
(a) The domain of h is all x for which is defined: (b) A little Chain Rule:
(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.
The standard description of the FTC is that “The two central operations of calculus, differentiation and integration, are inverses of each other.” —Wikipedia • A more useful description is that the two definitions of the definite integral: • The difference of the values of an anti-derivative taken at the endpoints, [definition used by Granville (1941) and earlier authors] • The limit of a Riemann sum, [definition used by Courant (1931) and later authors] • yield the same value.
2004 AB3(d) A particle moves along the y-axis so that its velocity v at time t ≥ 0 is given by v(t) = 1 – tan–1(et). At time t = 0, the particle is at y = –1. Find the position of the particle at time t = 2. y '(t) = v(t) = 1 – tan–1(et) y(t) = ?
Velocity Time = Distance time velocity distance
Change in y-value equals Since we know that y(0) = –1: If we know an anti-derivative, we can use it to find the value of the definite integral.
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)
(a) r = 2 miles. (b) A = 2πrΔr
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).