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I tried to write the integral as limit of a sum which gives:$$\int_0^1f(r-1+x)dx=\lim_{n\to\infty}\frac{1}{n}[f(0)+f(\frac{1}{n})+f(\frac{2}{n})...+f(\frac{n-1}{n})]$$

which i wrote as: $$\lim_{n\to\infty}\frac{1}{n}\sum_{r=0}^{n-1}f(\frac{r}{n})$$

and when I put this in the question, I get a double summation. I don't know how to go further. Kindly suggest.

Btw the answer to this question is $$\int_0^nf(x)dx$$

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For $\displaystyle\int_0^1 f(r-1+x)dx$ with substitution $u=r-1+x$ we have $$\int_0^1 f(r-1+x)dx=\int_{r-1}^r f(u)du$$ so $$\sum_1^n\int_0^1 f(r-1+x)dx=\int_{0}^n f(x)dx$$

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  • $\begingroup$ Thanks! Btw, is there any way this can be solved as limit of a sum? I am 100% fine with the answer you wrote but this question was given in definite integral as the limit of a sum exercise so just wondering if there's any? $\endgroup$ – Cheapstrike Jan 16 '17 at 6:37
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    $\begingroup$ @Cheapstrike For summation we must have $\Delta x=\Big((r+1)-1+x\Big)-(r-1+x)=r\to0$ but It does not occur. $\endgroup$ – Nosrati Jan 16 '17 at 6:40
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As a hint: try to substitute $u=r+x-1$.

(This would typically be a comment, but I do not, alas, have enough reputation...)

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  • $\begingroup$ I think you do now. Welcome! $\endgroup$ – numbermaniac Nov 25 '17 at 12:44
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$\int _0^1 f (r+x-1)=\int _0^1 f (r-x) $ let $r-x=u $ thus $dx=-du $ hence we have $I=\sum _1 ^n -\int _r ^{r-1} f (u)du =\sum _1 ^n \int _{r-1} ^r f (u)du =\int _0 ^n f (x)dx $

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  • $\begingroup$ from Second last to last step, is it a formula? $\endgroup$ – Kislay Tripathi Nov 25 '17 at 9:57
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    $\begingroup$ The disappereance of - sign? $\endgroup$ – Archis Welankar Nov 25 '17 at 9:58
  • $\begingroup$ Its a known fact that $\int _a^b f (x)dx=-\int _b ^a f (x)dx $ . I hope its clear now. $\endgroup$ – Archis Welankar Nov 25 '17 at 10:04
  • $\begingroup$ please tell me about The disappereance of summation sign? $\endgroup$ – Kislay Tripathi Nov 25 '17 at 10:06
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    $\begingroup$ I think you should read basic theorems of integration you can then answer your question yourself;) $\endgroup$ – Archis Welankar Nov 25 '17 at 10:08
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Let $\displaystyle \mathfrak{I}:=\sum_{r=1}^n \int_0^1 f(r+x-1) \,\text{d} x$. Leting $u:=r+x-1$ we have $\dfrac{\text{d}u}{\text{d}x}=1 $ . If $x=0$ we have $u=r-1 $, and if $x=1$ we have $u=r$. Therefore:

$$\begin{align}\mathfrak{I}=&\sum_{r=1}^n \int_{r-1}^r f(u) \, \text{d}u\\ \end{align}$$ But we know that $\displaystyle \int_{a}^b f \,+ \int_b^c f = \int_a^c f $. Therefore : $$\boxed{\,\,\,\mathfrak{I} =\int_{0}^n f(x) \, \text{d}x\,\,\,\,\,} $$

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  • $\begingroup$ Brother how did you conclude $\displaystyle \int_{a}^b f \,+ \int_b^c f = \int_a^c f$ From this $\begin{align}\mathfrak{I}=&\sum_{r=1}^n \int_{r-1}^r f (u) \, \text{d}u\\ \end{align}$ .Is it some kind of formula or i need knowledge of riemann integral for this? $\endgroup$ – Kislay Tripathi Nov 25 '17 at 11:43
  • $\begingroup$ @KislayTripathi it's the sum of areas. Think about the integrals as areas under a graph - if you add the area from $a$ to $b$ and the area from $b$ to $c$ together, wouldn't it be the same as the area from $a$ to $c$? $\endgroup$ – numbermaniac Nov 25 '17 at 12:17
  • $\begingroup$ Is a way to "glue" togheter the areas. Consider a function $f$ and $a<b<c$. The $\displaystyle \int_a^c f $ (area from $a$ to $c$) can be cuted in a certain $b$. Therefore the area will bem the sum of those areas, wich are $\displaystyle \int_a^b f $ (area from $a$ to $b$) and $\displaystyle \int_b^c f$ (area from $b$ to $c$). Literally a slice haha. $\endgroup$ – Gustavo Mezzovilla Nov 25 '17 at 13:46

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