AP Calculus

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Special Focus: The Fundamental Theorem of Calculus Teacher Notes, Answers, and Extensions The version of the Fundamental Theorem covered by this activity says that if f is a continuous function on the closed interval [a,b], then the function F( x) = ∫ f ( t) dt is a d x an antiderivative for f. Other ways to say this are F′ ( x) = f ( x), or f t dt f x dx ∫ ( ) = ( ) . a This is an important result in the history of the calculus, and one goal of this activity is to help students discover and appreciate its beauty. Students need to remain focused while carrying out the activity and think about each step. Class Management This activity takes at least one full class period (not counting the extension) to complete (45 to 50 minutes). If one class period is insufficient, students could complete the activity as a homework assignment. However, they should have at least created the lists and completed the first six questions in class. Teachers should work through the activity once before doing it with students. Depending on the calculator skills of your students, they could work independently, in groups, or with the teacher guiding the activity on an overhead display. Calculus Prerequisites This activity can be done any time after students have learned about a function defined by x a definite integral, like F( x) = ∫ g( t) dt. Students should know that such functions will a be positive when the integrand is a positive function and x > a, and negative when the integrand is a negative function and x > a. Although not essential, it is helpful if students know the addition property: Calculator Prerequisites b a c b ∫ ∫ ∫ c a f ( t) dt + f ( t) dt = f ( t) dt This activity makes extensive use of calculator functionality. Students must be able to define and graph functions and scatterplots, set up a viewing window, and find the zeros of a function on the graph screen. It introduces the calculator’s integral function and shows how to create a list using the seq command. TI-83 users need to know how to enter the name of a function (retrieving Y1, for example, from the VARS menu) and the name of a list (entering L1 for example, by typing 2nd 1). x AP® Calculus: 2006–2007 Workshop Materials 69
Special Focus: The Fundamental Theorem of Calculus In the TI-83 version of the activity, an additional argument, 0.1, is supplied to the fnInt command when defining the function Y3. This argument specifies the guaranteed precision of the value returned by fnInt, and using 0.1 speeds up the plotting somewhat. Preliminary Comments This exploration gives students hands-on experience with functions of the form x F( x) = ∫ f ( t) dt and emphasizes the connection between the function, F, and the 0 integrand, f. That connection is at the heart of the Fundamental Theorem. The hardest calculator work occurs before any questions are asked. If students get confused about what L3 represents, it may be helpful to have them evaluate fnInt(Y1(X),X,0,0.1), Y2(0.1), and L3(2)on the HOME screen. All three of those expressions are equal. The activity could be done using the calculator’s built-in TABLE feature and by graphing Y2 and Y3 directly, instead of as scatterplots. However, it takes a long time to redraw the graphs or redisplay the tables for functions defined by integrals. Since students are going back and forth between the graph and the table several times in the activity, the values of the integral are stashed away in lists. This process has a side benefit. It allows you to do calculations with those values. In particular, you can approximate the derivative of Y2 by forming difference quotients. A scatterplot of those approximations can then be compared with the graph of Y1. This extension appears after the answers and supports the conclusion students make at step 15. Answers and Additional Comments In the first six questions, students should begin to see that the function, Y2, is increasing wherever the integrand is positive and decreasing wherever the integrand is negative. This connection should resonate with students’ experience with increasing/decreasing behavior of a function and the sign of its derivative. 0 1. The first value in L3 represents ∫ Y1( T ) dT and so is 0. There is no area from 0 0 to 0. 2. The second value in L3 is positive since the integrand is positive for all T in the interval [0,0.1]. 70 AP® Calculus: 2006–2007 Workshop Materials
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AP Calculus

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