In class I stated that any 2nd or 3 rd order polynomial that goes through the points 0
1 ,
3 y
 
 
  and
1
1 ,
3 y
 
 
  would have the same integral over the range -1 to 1 as the line (remember, there is
only one) that goes through the same two points. The easiest way to prove that the areas are all
the same (and demonstrate why) is using NDDP.
First, let’s construct a 1st order NDDP through the points above, and calculate the area under that
line for -1 ≤ x ≤ 1. The NDDP equation for the 1 st order polynomial (straight line) that goes
through points  0 0,x y and  1 1,x y is:
       1
0 0 1 0 0 1 0,f x f x x f x x a a x x     
It will turn out that we don’t even need to know 0f or  1 0,f x x . They are simply constant
values that we can carry around. I’m going to replace them with simpler terms:
   1
0 1 0f x a a x x  
Now let’s integrate this line from -1 to 1:
    1
21
0
0 1 0 0 1
1 1
2
line
x x
A a a x x dx a x a
 
 
          

From the 2-point Gauss Legendre, 0 1 3
x  . Then
 
1
2
1
0
1
2 2
1
0
2 2
1 1
0 0
0 1
1 32
1 11 1 1 1
3 32
1 3 1 3 1 2 3 3 1 2 3 3
2 2
2 2 3 33 3
2
2 3
line
a
A a x x
a
a
a a
a a
a a

        
    
                       
           
           
       
 
Just to recap, this is the area under the curve, between x = -1 and x = +1, of the straight line that
goes through points at  0 0,x y and  1 1,x y . In truth, though, this value doesn’t matter. What
we do next is what matters.
The NDDP equation for the 2nd order (quadratic) polynomial that goes through points  0 0,x y ,
 1 1,x y , and  2 2,x y is:
          
    
2
0 0 1 0 0 1 2 1 0
0 1 0 2 0 1
, , ,f x f x x f x x x x x x f x x x
a a x x a x x x x
     
     
Now let’s integrate this quadratic polynomial from -1 to 1:
    
    
  
1
0 1 0 2 0 1
1
1 1
0 1 0 2 0 1
1 1
1
2 0 1
1
quad
line
A a a x x a x x x x dx
a a x x dx a x x x x dx
A a x x x x dx

 

       
       
   

 

I broke it into two integrals to help you visualize that, to make Aquad = Aline, what has to happen is
for the integral of the last term of the NDDP to be zero. Let’s see what it is:
  
1 1
2 0 1 2
1 1
1 1
3 3diffA a x x x x dx a x x dx
 
             
    
 
   
1 31 3 3
2
2 2 2
1 1
11 1 11 0
3 3 3 3 3 3 3
diff
x x
A a x dx a a
 
        
                        

Since we did not define the values of  2 2,x y , the 2 nd order NDDP expression above describes
every quadratic that goes through points  0 0,x y and  1 1,x y . The above identity proves that
the area under the curve from x = -1 to x = +1 for every quadratic that goes through points
 0 0,x y and  1 1,x y is equal to the area under the line through the same points and over the
same range.
To prove that the area under every cubic polynomial is the same as the area under the line, let’s
add another generic point  3 3,x y and look at the 3rd order NDDP.
          
     
        
0 0 1 0 0 1 2 1 03
0 1 2 3 2 1 0
0 1 0 2 0 1 3 0 1 2
, , ,
, , ,
f x x f x x x x x x f x x x
f x x x x x x x f x x x x
a a x x a x x x x a x x x x x x
      
  
     
         
The area under this curve, between x = -1 and x = +1, is:
        
    
   
1
0 1 0 2 0 1 3 0 1 2
1
1 1
0 1 0 2 0 1
1 1
1
3 0 1 2
1
cubicA a a x x a x x x x a x x x x x x dx
a a x x dx a x x x x dx
a x x x x x x dx

 

           
       
   

 

The first integral is the area under the line, and we already know that the second integral is zero.
(a) To show that the area under the cubic polynomial is equal to the area under the straight
line, show that the third integral in the above expression is zero
(b) To show that this only works for polynomials up to 3 rd order, add another generic point
 4 4,x y , express the 4th order NDDP, and integrate it from x = -1 to x = +1. We already
know that the integral of the first two terms of the NDDP is the area under the line, and
the integrals of the 3rd and 4th terms are zero. Show that the integral of the 5 th term in the
NDDP is not zero.