Unformatted text preview: S ) ≥ sup( { 1 21 2 n  n ∈ N } ) = 1 2 . Likewise we ﬁnd 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 ﬁnd 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
 323
 Tn, Equals sign, 1 k, 0 K, Step function

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