(New page: 2) Pick a partition P of [a,b] <math>a=x_0<x_1<...<x_n=b</math> Pick <math>c_i \in (x_0, x_1) \ i=1,...,N.</math> Let n=N Define <math>c_{n+1}=b</math> and <math>c_0=a</math>. Then ...) |
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<math>a=x_0<x_1<...<x_n=b</math> | <math>a=x_0<x_1<...<x_n=b</math> | ||
| − | Pick <math>c_i \in (x_0, x_1) \ i=1,..., | + | Pick <math>c_i \in (x_0, x_1) \ i=1,...,n.</math> |
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Define <math>c_{n+1}=b</math> and <math>c_0=a</math>. | Define <math>c_{n+1}=b</math> and <math>c_0=a</math>. | ||
| − | Then <math>\sum_{i=1}^n \phi(c_i)(f(x_i)-f(x_{i-1})) = \sum_{i=1}^ | + | Then <math>\sum_{i=1}^n \phi(c_i)(f(x_i)-f(x_{i-1})) = \sum_{i=1}^n \phi(c_i)f(x_i)-\sum_{i=1}^n \phi(c_i)f(x_{i-1}) |
</math> | </math> | ||
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<math> | <math> | ||
| − | = - f(b)\phi(b) + f(a)\phi(a) - \sum_{i=0}^{n} (\phi(c_{i+1})-\phi(c_{i})f(x_i)</math>. | + | = - f(b)\phi(b) + f(a)\phi(a) - \sum_{i=0}^{n} (\phi(c_{i+1})-\phi(c_{i}))f(x_i)</math>. |
Revision as of 16:03, 22 July 2008
2)
Pick a partition P of [a,b]
$ a=x_0<x_1<...<x_n=b $
Pick $ c_i \in (x_0, x_1) \ i=1,...,n. $
Define $ c_{n+1}=b $ and $ c_0=a $.
Then $ \sum_{i=1}^n \phi(c_i)(f(x_i)-f(x_{i-1})) = \sum_{i=1}^n \phi(c_i)f(x_i)-\sum_{i=1}^n \phi(c_i)f(x_{i-1}) $
$ = \phi(c_n)f(x_n)-\phi(c_1)f(x_0) + \sum_{i=1}^{n-1} (\phi(c_i)-\phi(c_{i+1})f(x_i) $
$ = - f(b)\phi(b) + f(a)\phi(a) - \sum_{i=0}^{n} (\phi(c_{i+1})-\phi(c_{i}))f(x_i) $.
(by adding and subtracting $ f(b)\phi(b) $ and $ f(a)\phi(a) $.
Taking the limit on both sides as $ |P| \rightarrow 0 $ gives us the claim.
--Wardbc 15:57, 22 July 2008 (EDT)
