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Im trying to integrate this, using theorem 7.9 of apostol's book:

$$\int^{10}_0 f(x)d\alpha(x) $$

$f(x) = x^2$ and $\alpha(x)= 3\chi(7,9](x)$ Where $\chi(x)$ is $0$ everywhere except $1$ in the interval $(7,9]$

So using this theorem, and knowing the steps are at 7, and 9, and everywhere else the sum and substraction of $\alpha$ is $0$ i get:

$$f(7)[\alpha(7+)-\alpha(7-)] + f(9)[\alpha(9+)-\alpha(9-)] = $$ $$= 7^2 (3*(1 - 0)) + 9^2 (3*(0-1)) = 3*7^2 - 3*9^2$$

Does this sound about right? In class we did a mostly theeoretical stuff with the R-S integral but not a single example so I'm i little lost.

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It seems you lost a factor of $3$ from $\alpha$. Otherwise, it's fine. –  Daniel Fischer Jul 12 '13 at 0:17
    
Here is a useful result. –  Mhenni Benghorbal Jul 12 '13 at 0:43
    
math.ubc.ca/~feldman/m321/step.pdf I think the above proof is a general proof for such problems. good luck –  user113458 Feb 2 at 4:33
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