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Unformatted text preview: S ) sup( { 1 21 2 n  n N } ) = 1 2 . Likewise we nd for the upper integral that { 1 2 + 1 2 n  n N } T = { R 1 t ( x ) dx  t ( x ) f ( x ) ,t ( x ) a step function } , so I = inf( T ) inf( { 1 2 + 1 2 n  n N } ) = 1 2 . Using Theorem 1.9 we see that 1 2 I ( x ) I ( x ) 1 2 . Thus we nd that all (weak) inequalities actually have to be equalities, so in particular 1 2 = I ( x ) = I ( x ) = 1 2 . The same theorem 1.9 now tells us that R 1 xdx = 1 2 . 1...
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 Spring '10
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