• Reimann Sums and the Integral

    
    Date: 01/18/99 at 03:34:41
    From: Brian J.
    Subject: Fund. Calc Theorem/Reinmann Sums
    
    Dear Dr. Math,
    
    I have always wondered why the limit of Reimann sums just happened to 
    be the integral? How is that derived? I guess a related question is 
    the proof of the fundamental theorem of calculus:
    
         /b
        |   
        |    =    F(b) - F(a)
       /a
             
    One source I had said that the proof involves advanced calculus, so it 
    wasn't included, but my textbook has a proof printed. Is it that 
    advanced? Thanks.
    

  • 
    Date: 01/18/99 at 11:03:48
    From: Doctor Anthony
    Subject: Re: Fund. Calc Theorem/Reinmann Sums
    
    If you draw the curve and consider an element of area dA formed by 
    verticals through the points x and x+dx along the x axis, then by 
    elementary geometry (and for a very short interval dx)
    
         dA = f(x) dx    where f(x) is the ordinate at point x
    
    So   dA/dx = f(x)
    
    Therefore f(x) is the derivative of the area A. Since by definition, 
    integration is the inverse of differentiation, the integral of f(x) 
    will be the area.
    
    That means
    
       F'(x) = f(x)
    
       F(x) = INT[f(x) dx]
    
    and on the interval [a,b] this gives
    
       INT[f(x) dx] = F(b) - F(a)
    
    so that F(b) - F(a) is the area under the curve between x = a and 
    x = b.
    
    Riemann sums, which calculate the areas algebraically (with n -> 
    infinity), will therefore approach the value of the integral on the 
    closed interval [a,b]. 
    
    The advanced calculus (analysis) is mostly in the rather long-winded 
    justification for algebraic manipulation of infinitesmals. You can 
    probably leave that to analysts and agree that the basic concept of 
    Riemann sums is very reasonable.
    
    - Doctor Anthony, The Math Forum