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I tried a thought experiment, just learning about calculus new to it in general and could not find this definition of right Riemann sum. This is the formula I created and it seems to work. [a,b] is interval you want to integrate from. n is the number of rectangles you want to create

$$ \Delta x \cdot \sum_{i=1}^n f(a+i \Delta x) $$

$$ \Delta x = \frac{(b-a)}{n} $$

Is this formula correct for getting the correct right riemann sum approximation?

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    Yes. https://en.wikipedia.org/wiki/Riemann_sum#Right_Riemann_Sum. – 2017-01-26

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Simple answer here! *drum roll please... *

YES! This becomes the definition of the Riemann Integral if you add $\lim_{n \to \infty}$ before the sum.

Disclaimer: if you gave any further questions on this feel free to ask in a comment

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    No, the Riemann sum is not the limit. The limit is the Riemann integral. – 2017-01-26
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    (Please do not add that disclaimer to your answers: there is no need, as that is how the *whole site* works…) – 2017-01-26
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    @MarianoSuárez-Álvarez eh, semantics. I agree with you though, I'll change it. – 2017-01-26
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    It is not semantics. The sum and the integral are very different things. Moreover, the question is precisely about the definition of something! – 2017-01-26
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    @MarianoSuárez-Álvarez the disclaimer was a joke ;) I'm well aware of how this site works – 2017-01-26
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You are right, this is indeed the definition of a right Riemann sum $R_n$ such that, $$R_n = \sum_{i=0}^n f(x_i)\Delta x$$ where $$x_i=a+i\Delta x$$ and $$\Delta x = \frac {b-a}n$$in the limit, as $n\rightarrow \infty $, the Riemann sum $R_n$ becomes the definite integral over the interval $[a,b]$, $$\lim_{n\to \infty} R_n= \lim_{n\to \infty} \sum_{i=0}^n f(x_i)\Delta x = \int_a^bf(x)dx$$ Since $\Delta x = \frac {b-a}n$, $\Delta x\rightarrow 0$ as $n\rightarrow \infty $ and therefore it is equivalent to say $$\lim_{\Delta x\to 0} R_n= \lim_{\Delta x\to 0} \sum_{i=0}^n f(x_i)\Delta x = \int_a^bf(x)dx$$ Hope this helps.