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2017 HAWAII UNIVERSITY INTERNATIONAL CONFERENCES SCIENCE, TECHNOLOGY & ENGINEERING, ARTS, MATHEMATICS & EDUCATION JUNE 8 - 10, 2017 HAWAII PRINCE HOTEL WAIKIKI, HONOLULU, HAWAII

AN ANALYSIS OF CAPITAL F IN THE FUNDAMENTAL THEOREM OF

GARRISON, JOSEPH M. DEPARTMENT OF MATHEMATICS MIDDLE GEORGIA UNIVERSITY COCHRAN, GEORGIA. Prof. Joseph M Garrison Department of Mathematics Middle Georgia State University Cochran, Georgia.

An Analysis of Capital F in the Fundamental Theorem of Calculus

Synopsis:

This paper will reflect on the integrating central to the statement of the theorem which will be called Capital F. Capital F’s relationship to all of the other antiderivatives of f(x) and the various structures of capital F producing the same function will be examined. An analysis of Capital F in the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus

x Suppose ftdt exists for all xc ,  ab , ; c is a real number. c

x Then if Fc xftdt  , it follows that c ddx Fc  x ftdtfx   x , c  ab , ; c is a constant real number. dx dx c After eluding Archimedes, Descartes, Barrow, Fermat and others (https://en.wikipedia.org/wiki/History_of_calculus) the fundamental theorem of calculus was finally discovered and validated through differentials and by and later rigorously proved using limits by Augustin Louis Cauchy (1789‐‐1857). “The modern proof of the Fundamental Theorem of Calculus was written in his Lessons Given at the École Royale Polytechnique on the Calculus in 1823. Cauchy's proof finally rigorously and elegantly united the two major branches of calculus (differential and ) into one structure.” https://www3.nd.edu/~apilking/Math10550/Lectures/26.%20Fundamental%20Theorem.pdf

This paper will reflect on the integrating function, which is central to the statement of the fundamental theorem of calculus. We will call this function Capital F. Capital F’s relationship to all of the other antiderivatives of f(x) and the various structures of capital F producing the same function will be examined.

Proposition One:

In other words, can all the antiderivatives of f(x) be generated by an appropriate choice of c in

x ftdt( ) where x,c ? c

This paper will show that in many cases it is not possible to generate all the antiderivatives of f(x) by an appropriate choice of c. Note that we are not arguing that a given antiderivative does not exist, but that it cannot be generated by choosing a real number for the lower index

x of the given Fxc  ftdt  which we are calling Capital F. c

Proposition Two: x Suppose Fxc  ftdt  and fx ( ) is continuous on ; c is a constant. c

Does there exist c1 c such that xx Fcc x ftdtF   x ftdt  ? cc1 1

In other words, are there different values of c, which produce the same F? This paper will show that quite often there are many different values of c producing the same Capital F. These values occur at the real zeros of Capital F!

Proposition One analyzed, first glance: Can all the antiderivatives of f x be generated by an appropriate choice of c in x ftdt() where x,c ? c x Let Fxc  ftdt  for a given c . c Now consider an antiderivative of fx , say Gx  , obtained by choosing an appropriate k11 so that Gx Fc  x k and Gx has no real roots. This k1 exists when Fc  x has either an upper bound or a lower bound. Without loss of generality assume l is a lower bound.

Choose kl1  .

Since Fxc  l ,

Fxcc k11 l l00 Fx  k . Then G(x) cannot be represented by x Gcc x ftdt  ,c222 because G  c  0 c is a real root. 2 c 2 2 But G(x) has no real root on  .

Analysis Continued: x Let Fx ( ) tdt be a given antiderivative of fx ( ) x . 0 0 x x 111222 Fx00() x because Fx () tdt t x 2220 0 (fundamental theorem of calculus). Now any other antiderivative, say G(x), can be 1 represented by Gx ( ) F ( x ) k x2 k ( ). 0 2 x Now what value of c in Fx ( )  tdt c c 1 would produce the xk2  ? 2 111x Since xkFx222( )  tdtx   c , 222c c 1 then ckc222 k 2 k 0. Therefore no antiderivative 1 of the form yxkk2 ;  2 x can be represented by tdt . c

Example for proposition One: Suppose f (xx ) .

1 2 Gx( ) y2 x  4 is a specific antiderivative for f ( x ). x Let Fx( )  tdt c What real value for c above would generate Gx ( )? x Fx( ) tdt is one antiderivative (fundamental theorem of calculus) c Any other antiderivative of x can be represented by

1 2 2 xk where k is a constant real number(mean value theorem). x x 11112222 Fx() tdt 2222 t  x c x 4 c c The question becomes whether there exists c such that

1 22  2 c 4 which would require that c 8 and then c . 1 2 Any antiderivative of the form 2 xk x must have k 0 to be represented by Fx ( ) tdt . c

1 2 Therefore, any antiderivative of the form 2 x  k where k is positive

x cannot be represented by Fx() tdt. c x Let Fx ( ) ftdt ( ) be a given antiderivative of fx ( ) that is continuous on . ci c i x Fx( ) ftdt ( ) exists by the fundamental theorem of calculus. ci c i Now any other antiderivative, say G(x), can be x represented by Gx ( ) F ( x ) k ftdt ( ) +k (mean value theorem). ci c i x Fx() ftdtFx () () Fc ( ). ccciiiic i x Now what value of c in Fx ( ) ftdt ( ) c c would produce the antiderivative Gx ( ) F ( x ) k ? ci x Gx() F () x k ftdt () Fx() Fc () ci c ccii Fc() k ci and Fc( ) k 0 Gc ( ) ci  that for the antiderivative Gx ( ) F ( x ) k ci x to be represented by ftdt ( ) , then the set containing the real roots c ofGx ( ) must not be empty; and furthermore all elements in this set produce x identical Fx ( ) ftdt ( ) . ci c i

More Examples for Proposition One:

x Antiderivatives with no real lower index in the generating function Fx() tdt : c 1 Fx() x2 1 1 2 1 Fx() x2 4 2 2 1 Fx() x2 16 3 2

Example:

x Antiderivatives with real lower index in the generating function Fx() tdt : c 1 Fx() x2 when c=0 1 2 1 Fx( ) x2 2 when c 2 2 2 1 Fx( ) x2 8 when c 4 3 2 1 Fx() x2 18 when c 6 4 2 1 Fx() x2 32 when c 8 5 2

Proposition Two analyzed: xx

Suppose c1 c and f ( tdt ) f ( tdt ) . cc 1 xx f() t dt f () t dt 0 cc 1 xc ftdt()1 ftdt () 0 cx c 1 ftdt() 0 c x cftdt1 is a root of ( ) . c x We also know that c is a root of ftdt ( ) ; c therefore the antiderivatives are = when the c1 's are roots.

x Suppose Fxc  ftdt  . 1 c 1 Suppose c is a root of Fxcc ,  212c1 note that c is a also root of Fx . 1 c1

c1 Then ftdt  0 c 1

c2 and Fcc 22 ftdt  0 because c is a root. 1 c 1 Consider xcc112 the open  , . cxc212 f x dx f  x dx f  x dx 0 ccx111

xcx121 or fxdx fxdx   fxdx   cxc112 Fx() Fx (); cc1211 therefore Fx ( ) Fx ( ), and there are two distinct values cc12 of c producing the same Capital F.

Example for Proposition Two:

x Suppose Fx( ) (2 t 3) dt . 1 Then F(1) 0, and all antiderivatives are of the form y  xxc2  3 . Fcc(1) 0 1 3 2, so Fx( ) x2 3 x 2. Note that f(2)=0. xx Fx() (2 t 3) dt Fx () (2 t 3). dt 12 Thus, we have two values for the lower of integration producing the same F(x). These lower limits always occur at the real roots of F()xx2 3 x 2.

The Connection between Proposition I and Proposition II: Suppose there exists a real c such that xx Fx( ) ftdtFx ( ) . ( ) ftdt ( ) is an antiderivative of fx ( ), cccc continuous on  (fundamental theorem of calculus). Now any other antiderivative, say G(x), can be represented by Gx ( ) Fc ( x ) k (mean value theorem). x What value of c in Fx ( ) ftdt ( ) 1c1 c 1 would produce the antiderivative Gx ( ) Fc ( x ) k ? xx Gx() F () x k f ()tdtk F () x f ()tdt  F () x F ( c ) ccc1cc1 c 1

Fxcc() Fc () k Fxc1() Fcc ( )

Fxccc1() 0 k Fx ()  Fc ( ) and Fcc1 ( ) k .

Thus for the antiderivative Gx ( ) Fc ( x ) k x to be represented by ftdt ( ) , the set containing the roots c 1 ofFcc1 ( ) k , or Gx ( )  0 must not be empty; and furthermore, all elements in this set are real numbers producing x identical functions Fx ( ) ftdt ( ) . c c

x Given an antiderivative, Fx() ftdt () ,of a and a desired constant c c of integration, k, in another antiderivative, in order to determine if the new antiderivative can x be written in terms of Capital F with a real c solve the Fx() k ftdt () =0 c c 1

or solve Gx ( ) 0.

If there are real roots, it can be written in terms of Capital F with a real c. The real roots all produce equal Capital F’s.

Example:

x Suppose Fxc ( ) 2 t 1 dt . c x Fx0 ()21tdt 0 22x tt0 xx(fundamental theorem). Analyze xx2 6 0 The roots  3, 2 are values for c that x 2 produce the same Fxc ( ) 2 t 1 dtx =  x 6. c

xx 2 Fx2, 3() 21 t dtt  21 dtxx 6 23

Example: x Suppose Fxc ( ) e k . x xt What values of k allow Fc (xek ) to be written as edt ? c Then ekx 0 when k 0.

Concluding Example

x Note that any of odd degree can be represented by f ()tdt. c

Suppose G(x)xxx32 2  2. x What values of c allow G(x) to be written as F (xttdt ) 32 4 1 ? c Solve G(x)xxx32 2  2 0. The roots are  1, 1, 2 . G(x)xxx32 2  2 x 341ttdt2   1 x 341ttdt2   1 x 341.ttdt2  2

Conclusion: Let f(x) be the derivative of a , say G(x). The integrating functions x F ()xftdt () used in the Fundamental Theorem ci c i are equal when the csi ' are real roots of G(x), and x F ()xftdtGx () (). There are many times when ci c i an antiderivative of f(x) cannot be generated by x cFxftdt in ( )  ( ) . This occurs when G(x) has i c i no real root. The lower index ci determines the value of the k, in the antiderivative Fx() Gx () k where Gx()is any antiderivative because ci is a root ofFx(). The final concern is whether there is a representation of G(x) x =F ()xftdt () where f(x) is not the derivative of ci c i G(x). The fundamental theorem of calculus guarantees that no such function exists. QED