On-line Math 21

5.1  The definition of the Riemann Integral

Proposition 1 Rules

1. 
   

   ó õ b a  c dx = c(b-a).

2. 
   

   ó õ b a  f(x)dx = - ó õ a b  f(x)dx.

   Actually, this rule is a definition. We hadn't had a definition for the right-hand side (at least when a < b ); this helps for some later formulas.
3. 
   

   ó õ a a  f(x)dx = 0.

   As before, this is really a definition.
4. 
   

   ó õ b a  cf(x)dx = c ó õ b a  f(x)dx.

5. 
   

   ó õ b a  f(x)±g(x)dx = ó õ b a  f(x)dx± ó õ b a  g(x)dx.

6. If f(x) ³ 0 for all x Î [a,b] , a < b , then
   

   ó õ b a  f(x)dx ³ 0.

7. If f(x) ³ g(x) for all x Î [a,b] , a < b , then
   

   ó õ b a  f(x)dx ³ ó õ b a  g(x)dx.

8. For all a , b , and c ,
   

   ó õ c a  f(x)dx+ ó õ b c  f(x)dx = ó õ b a  f(x)dx.

9. If m £ f(x) £ M for all x Î [a,b] , a < b , then
   

   m(b-a) £ ó õ b a  f(x)dx £ M(b-a).

10. 
    

    ê ê ó õ b a  f(x)dx ê ê £ ó õ b a  | f(x)| dx.

Rule #1 is pretty obvious. Any Riemann sum for f(x) = c a constant will have the same total, c times the length of the interval. Rules #2 and #3 are, as mentioned, really definitions. They will be useful for some of the later rules, though. Rule #4 is also clear, since the c can be factored out of any Riemann sum.

Rule #5 follows from the fact that, in any Riemann sum,

n å i = 1  f±g(xi*)Dxi = n å i = 1  ( f(xi*)±g(xi*)) Dxi = n å i = 1  f(xi*)Dxi+ n å i = 1  g(xi*)Dxi,

separating the Riemann sum into Riemann sums for f and g separately. Because of this, the same separation holds for the integrals.

Rule #6 is again obvious, since any Riemann sum would be positive, so must be the integral. This rule is, however, often forgotten in the heat of evaluating integrals using methods we'll deal with later on. This rule should be used as a rough check on other work: if the function you are integrating is positive, then the integral better be positive as well, or something went wrong.

Rule #7 follows from #6 and #5. Since f(x) ³ g(x) , then f(x)-g(x) ³ 0 , so

ó õ b a  f(x)-g(x) dx ³ 0.

But

0 £ ó õ b a  f(x)-g(x) dx = ó õ b a  f(x) dx- ó õ b a  g(x) dx,

so

ó õ b a  f(x) dx ³ ó õ b a  g(x) dx.

At the risk of repeating myself, again Rule #8 is obvious, at least if a £ b £ c , since a Riemann sum for

ó õ b a  f(x)dx

and for

ó õ c b  f(x)dx,

when added together, will give a Riemann sum for

ó õ c a  f(x)dx.

The result follows, in that case. If, on the other hand, the numbers a , b , and c are not in increasing order, all you have to do is re-arrange them in increasing order, making sure to change the sign of the integrals if you switch order. It will all balance out.

Rule #9 is just two applications of Rule #7.

Rule #10 is a bit cleverer. Since

-|f(x)| £ f(x) £ +|f(x)|,

two more applications of Rule #7 implies that

- ó õ b a  |f(x)|dx £ ó õ b a  f(x)dx £ + ó õ b a  |f(x)|dx,

which becomes Rule #10 when you turn that string of inequalities into an inquality with absolute values ( -A £ B £ AÛ |B| £ A ). This last rule is often very useful theoretically.

Copyright (c) 2000 by David L. Johnson.