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$ \int_{0}^2 x\,d \alpha $ where $ \alpha (x) = x $ if $ 0\le x\le 1 $ and $ \alpha(x)=2+x $ when $ 1<x\le 2 $ I did this by taking a partition which divided the interval $[0,2]$ to $2n$ equal parts

$P=\left\{{0,\frac{1}{n},\frac{2}{n},\dots,\frac{n}{n}=1,1+\frac{1}{n},\dots,2\cdot\frac{n}{n}=2}\right\}$

and considered a Riemann sum using $t_ i=x_i$ where $t_i \in $[$x_{k-1},x_k$] and by taking the limit when $n$ goes to infinity got $3$ as the answer. I wanted to check if its correct and if so why is it wrong to do it as below

$ \int_{0}^2 x d \alpha = \int_{0}^1 x d x + \int_{1}^2 x d (x+2) $ would get $2$ as the answer?

Thank You

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1 Answer 1

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The question is what happens at the "jump". In your case, the jump is at the point $1$, so $$ \int_{1}^2 x\; d (x+2) \ne \int_1^2 x\, d\alpha(x) $$

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  • $\begingroup$ It must change to equality by the theorem $$\int_a^bfdG=\int_a^bfG'dx$$ when $G$ is differentiable. $\endgroup$
    – Ali
    Aug 18, 2015 at 13:01
  • $\begingroup$ A function like $\alpha$, which is not continuous, is of course not differentiable at the point $1$. $\endgroup$
    – GEdgar
    Aug 18, 2015 at 16:31
  • $\begingroup$ What about $\alpha=I-x?$ $\endgroup$
    – Ali
    Aug 18, 2015 at 17:28

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