AP Calculus

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Special Focus: The Fundamental Theorem of Calculus Key Ideas That Help Students Understand the Fundamental Theorem Steve Olson Hingham High School Hingham, Massachusetts and Northeastern University Boston, Massachusetts I believe that we must focus on two important ideas as we help our students learn about the Fundamental Theorem of Calculus. 1. The evaluation part of the Fundamental Theorem can and should be introduced to students so they can see that the result is reasonable. Although students do not usually have difficulty in applying this part of the theorem, teachers can introduce this part in a way that helps students to believe that the result is almost “obvious.” 2. The antiderivative part of the Fundamental Theorem guarantees that a continuous function on an open interval has an antiderivative, whether or not a closed form of the antiderivative can be found. Often students view this part simply as a way to take the derivative of a function that is defined by a definite integral. Teachers must help students understand that x F( x) = ∫ f ( t) dt is the antiderivative of f whose graph contains the point (a, 0). a Thus, this part allows us to write the solution to an initial value problem when there is no known closed form antiderivative. The discussion that follows illustrates how I try to deal with these issues in my classroom. Evaluation Part of the Fundamental Theorem In an attempt to help students not only understand this part, but to help make this part seem obvious, I have broken the development into three steps. Evaluation Step 1: Understanding that partitioning a time interval and summing the products of the form (rate of change) × (subinterval width) for each subinterval gives an approximation for the total change AP® Calculus: 2006–2007 Workshop Materials 7
Special Focus: The Fundamental Theorem of Calculus I start by having my students complete the following exploration that uses a table of data comprised of nondecreasing speeds of a car at equally spaced moments in time to estimate the distance traveled. A car is traveling so that its speed is never decreasing during a 10-second interval. The speed at various moments in time is listed in the table below. Time in seconds 0 2 4 6 8 10 Speed in ft/sec 30 36 40 48 54 60 (a) Use the table to help explain why the best lower estimate for the distance traveled in the first 2 seconds is 60 feet. (b) Use the table to help explain why the best upper estimate for the distance traveled in the first 2 seconds is 72 feet. (c) Use the table to give the best lower estimate for the distance traveled in the first 10 seconds. An answer of 300 feet (which ignores some data in the table) is not correct. (d) Use the table to give the best upper estimate for the distance traveled in the first 10 seconds. These sums of products that you have found in (c) and (d) are called Riemann sums. (e) If you choose the lower estimate for your approximation of how far the car travels, what is the maximum amount your approximation could differ from the exact distance? (f) Choose speeds to correspond with t = 1, 3, 5, 7, 9. Keep the nondecreasing nature of the above table and do not select the average of the consecutive speeds. Find new best upper and lower estimates for the distance traveled for these 10 seconds. In parts (a) and (b), I am helping students enter the problem smoothly. If questions arise here, I can address them quickly. Most students will answer (a) by saying, “The minimum speed during the first 2 seconds is 30 ft/sec, so distance is at least (30)(2)=60 feet.” They will answer (b) by saying, “The maximum speed during the first 2 seconds is 36 ft/sec, so the distance is at most (36)(2)=72 feet.” 8 AP® Calculus: 2006–2007 Workshop Materials
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AP Calculus

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