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Special Focus: The Fundamental Theorem of Calculus Riemann wanted to know how discontinuous a function could be and still be integrable. To even begin to ask this question, he needed a very clear definition of what it means to be integrable. As he showed, his definition of integration allows for very strange functions. One of his examples is a function that makes a discontinuous jump at every rational number with an even denominator (every interval contains an infinite number of these), that is continuous at all other numbers, and that is integrable. 1 The ability to integrate discontinuous functions creates problems for the antiderivative d x part of the FTC: f t dt f x dx ∫ ( ) = ( ). Derivatives cannot have jumps. This statement a is deeper than the Intermediate Value Theorem. If a function has a derivative, then it must be continuous, but if a function is a derivative, it does not have to be continuous. Nevertheless, even discontinuous derivatives are almost continuous. If the limit from the left and the limit from the right at a exist, then they must be equal. If we let f be Riemann’s very discontinuous but still integrable function, then the derivative of x ∫ f ( t) dt cannot exist at rational numbers with even denominators. a d x If f is continuous at every point in ⎡⎣ a, b⎤ ⎦ , then it is true that f t dt f x dx ∫ ( ) = ( ) at every a x in ( a, b ) . The remainder of the nineteenth century would be spent figuring out what could be said about this antiderivative part of the FTC when f is not continuous at every point in ⎡⎣ a, b⎤ ⎦ . Beyond Riemann A derivative cannot have jumps, but it does not have to be continuous. One example is the derivative of ⎧ 2 ⎛ 1⎞ ⎪ x sin , x , F( x) = ⎝ ⎜ x ⎠ ⎟ ≠ 0 ⎨ ⎪ ⎩ 0, x = 0. As long as x is not zero, we can use the standard techniques of differentiation to find its derivative: ⎛ 1⎞ ′ = ⎝ ⎜ ⎠ ⎟ + 2 ⎛ −1⎞ ⎛ 1⎞ ⎛ 1⎞ F ( x) 2xsin x x ⎝ ⎜ ⎠ ⎟ cos 2 ⎝ ⎜ ⎠ ⎟ = x x ⎝ ⎜ ⎠ ⎟ − ⎛ 1⎞ 2xsin cos ⎝ ⎜ ⎠ ⎟ , x ≠ 0. x x AP® Calculus: 2006–2007 Workshop Materials 109
Special Focus: The Fundamental Theorem of Calculus When x is 0, we have to rely on the limit definition: h F h − F F′ ( ) = lim ( ) ( 0 0 ) = lim h→0 h h→0 2 ⎛ 1⎞ sin ⎝ ⎜ h⎠ ⎟ − 0 1 = h 0 h lim sin ⎛ ⎞ h 0 ⎝ ⎜ h⎠ ⎟ = . → This function is differentiable at all values of x, but the limits from the left and right of F′ ( x) as x approaches 0 do not exist. The derivative F′ ( x) oscillates forever between 1 and –1 as x approaches 0, and so F ′ is not continuous at x = 0. If we modify this function slightly: ⎧ 2 ⎛ 1 ⎞ ⎪ x sin , x , G( x) = ⎝ ⎜ x ⎠ ⎟ ≠ 0 2 ⎨ ⎪ ⎩ 0, x = 0, then we get a function that is still differentiable at every value of x, including x = 0, but the derivative is ⎧ ⎛ 1 ⎞ ′ = ⎝ ⎜ ⎠ ⎟ − −1 ⎛ 1 ⎞ ⎪ 2xsin 2x cos ⎝ ⎜ ⎠ ⎟ , x ≠ 0, G ( x) 2 2 ⎨ x x ⎪ ⎩ 0, x = 0. This derivative is not bounded near x = 0. For this function, the conclusion of the evaluation part of the FTC fails. Since G ′ is not bounded, the Riemann integral b ∫ G ′( x ) dx does not exist if 0 is in the interval ⎡⎣ a, b⎤ ⎦ , so it does not equal G( b) − G( a). a You can get around this difficulty for the function G by using improper integrals, only integrating up to a value ε near 0 and then taking the limit as ε approaches 0. But in 1881, Vito Volterra showed how to build a function (call it V) that is differentiable at every x and whose derivative stays bounded, but for which the derivative cannot be integrated. 2 Although the derivative stays bounded, the limit of the Riemann sums of the derivative does not exist, and we cannot get around the problem by using improper integrals. Volterra’s function is an even stranger example of a function for which the 1 You can find this function in my book, A Radical Approach to Real Analysis (Mathematical Association of America, 1994), pages 254–256. 110 AP® Calculus: 2006–2007 Workshop Materials
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