AP Calculus

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Special Focus: The Fundamental Theorem of Calculus x ∫ 2 Figure 1. F( x) = − dt t . Here we illustrate the case x = –5 with F(–5) = ln(5/2). As an illustration of these questions, consider Figure 1 above. That the input variable of the function F is located in the upper limit of integration is a stumbling block for students, so the graph can help students see that the variable x is sliding along the t- axis, and there is a definite domain (in this example, the domain of x is ( −∞, 0 ) ). As x varies in this example, the values of F(x) are 0 (at x = –2), negative (for − 2 < x < 0 ), and positive (for x < −2 ). Requiring f to be continuous helps students to recall the definition of the definite integral as a limit of Riemann sums, which are guaranteed to converge for continuous functions. A computer algebra system or a graphing calculator allows one to graph the function F(x) and then ask what properties of F(x) are apparent. The students should be able to determine from the graph that the function F(x) seems to be differentiable. We can prove the antiderivative part of the Fundamental Theorem from first principles using the definition of the derivative. This is a nice approach, since the properties of the definite integral are used to simplify the result below. For convenience, assume f ( x) > 0 on the open interval containing a. First notice that F x + h − F x F′ ( x) = lim ( ) ( ) = lim h→0 h h→0 ∫ x+ h a ∫ x f ( t) dt − f ( t) dt a = lim h h→0 ∫ x+ h x f ( t) dt . h To complete the proof, let h > 0 and bound the area representing ∫ f ( t) dt using x the minimum and maximum values of the function f on the domain [ x, x + h] , arriving at the inequality h ⋅ ⎛ min f t f t dt h ⎝ ⎜ ( ) ⎞ [ ] ⎠ ⎟ ≤ x h ∫ ( ) ≤ ⋅ max x ⎝ ⎜ + ⎛ t∈ x, x+ h t∈[ x, x+ h] x+ h ⎞ f ( t) . ⎠ ⎟ AP® Calculus: 2006–2007 Workshop Materials 99
Special Focus: The Fundamental Theorem of Calculus Figure 2. As h → 0 + , we see x + h → x, min Then ⎛ ⎝ ⎜ min ⎞ f ( t) ⎠ ⎟ ≤ ∫ t∈ [ x, x+ h] x+ h x f ( t) → f ( x), and max f ( t) dt h ≤ ⎛ ⎝ ⎜ max t∈ [ x, x+ h] t∈[ x, x+ h] ⎞ f ( t) . ⎠ ⎟ t∈ [ x, x+ h] f ( t) → f ( x). The limits of the left and right sides of the inequality are both equal to f ( x) as h → 0 + because f is continuous (see Figure 2), and so the Squeeze Theorem implies that F lim ( x + h ) − F ( x ) f ( t) dt x = lim = + h + h h→0 h→0 ∫ x+ h f ( x). For the limit as h → − 0 , we can write F lim ( x + h ) − F ( x ) F lim ( x ) − F ( x − k = ) = lim − h + k h→0 k→0 k→0 + ∫ x x−k f ( t) dt , k where you can replace h by −k with k > 0. Using the same idea as before, but with k instead of h, we have k ⋅ ⎛ min f t f t dt k ⎝ ⎜ ( ) ⎞ [ ] ⎠ ⎟ ≤ x ( ) ≤ ⋅ ⎛ max ∈ − x−k ⎝ ⎜ ∫ t x k, x t∈[ x−k, x] ⎞ f ( t) ⎠ ⎟ . Hence ⎛ ⎝ ⎜ min ⎞ f ( t) ⎠ ⎟ ≤ ∫ x x−k f ( t) dt k ≤ ⎛ ⎝ ⎜ max t∈[ x−k, x] t∈[ x−k, x] ⎞ f ( t) ⎠ ⎟ . 100 AP® Calculus: 2006–2007 Workshop Materials
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AP Calculus

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