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Calculus Challenge Problem #7 Solution Non-Fundamental Functions: When Wrong is Right! A student was wondering why I took off points for his solution to the following integration problem. 1 x 2 2 x 53 dx 12 2 1 53 02 2 0 53 3 . 0 First, note the error in the student’s work. I patiently showed him that the correct method was 03 x3 1 13 2 2 2 2 5 5 5 5 x 2 x dx x x 1 1 0 3 0 3 . 3 3 3 0 3 0 3 3 1 He, of course, argued that since his way was easier and got the same answer, it was a better solution. This got me wondering… 1 Are there other functions for which f x dx f 1 f 0 ? Since the fundamental theorem 0 of calculus is usually required to evaluate a definite integral, we will call a function for which b f x dx f b f a a non-fundamental function on [a, b]. We see that f x x2 2x 53 a is non-fundamental on [0, 1]. 1. Find a linear function that is non-fundamental on [0, 1]. 1 We need to find a function f x ax b so that ax b dx a b b a and 0 1 ax b dx 1 2 a b , so a 2b . Any function, ( y 6 x 3, y 2 x , etc) will work. The 0 basic function is f x kx 2. k . 2 Are there any other quadratic functions that are non-fundamental on [0, 1]? 1 Sure, lots of them. Now we need that ax 2 bx c dx a b c c a b and 0 1 ax 2 bx c dx 13 a 12 b c , so 13 a 12 b c a b . So, as long as c 23 a 12 b , the quadratic 0 will be non-fundamental. The simplest solution is to set b = 0, so f x kx 2 3. 2k 3 Can you find all polynomials that are non-fundamental on [0, 1]? Generalizing the previous result, we see that there is always a condition on the constant that will make the sum work. For example, with the 6th degree polynomial f x a0 a1x a2 x2 a3 x3 a4 x 4 a5 x5 a6 x6 , we require a1 2a2 3a3 4a4 5a5 6a6 . Clearly, we could just as easily choose any other 2 3 4 5 6 7 coefficient instead of the constant. Again, the simplest function satisfying these conditions is 6k n f x kx 6 . In general, the nth degree polynomial f x kx n k satisfies the 7 n 1 conditions. a0 4. Must the interval of integration be [0, 1] for polynomials to be non-fundamental? Can we find a more general result? No, [0, 1] is convenient, but any other interval will do. U [L, U]. Then ax n b dx L Consider f x ax n b on the interval a U n 1 Ln 1 b U L must equal a U n Ln . Solving n 1 a U n 1 Ln 1 b U L a U n Ln , so n 1 n 1 n 1 n 1 n 1 U n Ln U n Ln a U L a U L ba a . For every a, U, and L, U L n 1 U L U L n 1 U L there is a b that works. 5. Can functions other than polynomials be non-fundamental on some interval? If so, give some examples. The obvious choices are exponentials and sines and cosines. Karl Gross, o Bellville High School contributed the following example: e Consider f x ln 1 x c on [0, 1]. Then 2ln 2 1 c ln 2 ln1, so c 1 ln 2 ln . 2 e So, f x ln 1 x ln is non-fundamental on [0, 1]. But we know that the Maclaurin 2 expansion for f x ln 1 x is ln 1 x x a2 1 2 , a3 13 , …. We have from our earlier work that a0 x 2 x3 x 4 2 3 4 xn n with a0 0 , a1 1 , a1 2a2 3a3 4a4 5a5 6a6 makes the function 2 3 4 5 6 7 non-fundamental, so we must have 2 1 31 e 1 ln 1 3 2 43 2 2 n 1 n 1 n 1 n and we can use the non-fundamental property to find the sum of the infinite series. Reference: Graham, Jeffery A., Self-Integrating Polynomials, The College Mathematics Journal, September, 2005 as presented in The Calculus Collection edited by Caren Diefenderfer and Roger Nelson, MAA.