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intx - S ≥ sup 1 2-1 2 n | n ∈ N = 1 2 Likewise we...

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THE INTEGRAL R 1 0 xdx FOKKO VAN DE BULT We want to calculate the integral R 1 0 xdx , so we have f ( x ) = x and a = 0 and b = 1. Consider the step function s n ( x ) = b nx c /n . Then we clearly have s n ( x ) = b nx c /n ( nx ) /n = x = f ( x ). Let us moreover define t n ( x ) = d nx e /n , then t n ( x ) f ( x ). As partition for both s n ( x ) and t n ( x ) we can take x k = k n for 0 k n . Thus the value of s n ( x ) on ( x k - 1 , x k ) equals k - 1 n (as nx ( k - 1 , k ), so b nx c = k - 1 for those x ). The value of R 1 0 s n ( x ) dx is now equal to Z 1 0 s n ( x ) dx = n X k =1 k - 1 n ( x k - x k - 1 ) = n X k =1 k - 1 n · 1 n = 1 n 2 n X k =1 ( k - 1) = 1 n 2 n ( n - 1) 2 = 1 2 - 1 2 n . For the calculation of n k =1 ( k - 1) = n ( n - 1) 2 we used the famous identity which is given as exercises I.4.4.1a and I.4.7.5 in the book. Likewise we can calculate t n ( x ) = k for x ( x k - 1 , x k ) and Z 1 0 t n ( x ) dx = n X k =1 k n ( x k - x k - 1 ) = n X k =1 k n · 1 n = 1 n 2 n X k =1 k = 1 n 2 n ( n + 1) 2 = 1 2 + 1 2 n . Now we see that { 1 2 - 1 2 n | n N } ⊂ S = { R 1 0 s ( x ) dx | s ( x ) f ( x ) , s ( x ) a step function } , so I
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Unformatted text preview: S ) ≥ sup( { 1 2-1 2 n | n ∈ N } ) = 1 2 . Likewise we ﬁnd for the upper in-tegral 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 par-ticular 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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