Wikipedia:Reference desk/Archives/Mathematics/2011 November 30

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November 30

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Integral limit

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If f is continuous on [0,1], how to show that

 

I've been asked to compute the integral, which is easily done by choosing f = 0; but how could you show that it's true for all such f? Please assume only a knowledge of the fundamental theorem of calculus, if possible; I'd like to avoid log. —Anonymous DissidentTalk 12:29, 30 November 2011 (UTC)[reply]

You can use
 
So it's enough to calculate
 
You can either do this by L'Hopital's rule, or, if that's not allowed, write it as
 
By taking derivatives, show that the factor in parenthesis is increasing and bounded below, and therefore has a limit as  . Sławomir Biały (talk) 12:46, 30 November 2011 (UTC)[reply]
(e.c.) Just use the fact that a continuous function on a closed bounded interval is bounded: for some   there holds   for all  . So you can bound the absolute value of that expression replacing   by the constant  . In this case integrating then get  , that notoriously converges to   as  ; you conclude thanks to the sandwich theorem. --pma 12:53, 30 November 2011 (UTC)[reply]