Download Calculus Challenge #7 SOLUTION

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 
ba

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 31
e 1
ln    1        
3 2 43
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.