# What is a good technique for evaluating definite Riemann Integrals with Riemann Sums?

Although it is much easier by hand to work out definite Riemann integrals using the Fundamental Theorem of Calculus, I am interested improving my skill at evaluating Riemann integrals with Riemann sums. I'll give you an example of how I struggle with this technique.

[0]: For the unfamiliar, this is the right Riemann sum:

$R_{n}=lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i})\Delta x$

[1]: This is the integral I tried to evaluate:

$R_{n}=\int_{a}^{b} f(x) dx=\int_{1}^{2} x^3 dx$

[2]: To evaluate the sum, I need values of $\Delta x$, and $x_{i}$.

$\Delta x=\frac{(b-a)}{n}=\frac{(2-1)}{n}=\frac{1}{n}$

$x_{i}=a+i\Delta x=1+\frac{i}{n}$

[3]: Putting [0], [1], and [2] together:

$R_{n}=\int_{1}^{2} x^3 dx = lim_{n\to\infty}\sum_{i=1}^{n}(1+\frac{i}{n})^3\frac{1}{n}$

[4]: From here on, I am hoping for suggestions. This is how I started evaluating the sum: $R_{n}=lim_{n\to\infty}\sum_{i=1}^{n}(1+\frac{3i}{n}+\frac{3i^{2}}{n^{2}}+\frac{i^{3}}{n^{3}})\frac{1}{n}$

$R_{n}=lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}(1+\frac{3i}{n}+\frac{3i^{2}}{n^{2}}+\frac{i^{3}}{n^{3}})$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[\sum_{i=1}^{n}(1) + \sum_{i=1}^{n}(\frac{3i}{n}+\frac{3i^{2}}{n^{2}}+\frac{i^{3}}{n^{3}})]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \sum_{i=1}^{n}(\frac{3i}{n}+\frac{3i^{2}}{n^{2}}+\frac{i^{3}}{n^{3}})]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}\sum_{i=1}^{n}(3+\frac{3i}{n}+\frac{i^{2}}{n^{2}})]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}(\sum_{i=1}^{n}(3)+\sum_{i=1}^{n}(\frac{3i}{n}+\frac{i^{2}}{n^{2}}))]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}(3n+\sum_{i=1}^{n}(\frac{3i}{n}+\frac{i^{2}}{n^{2}}))]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}(3n+\frac{i}{n}[\sum_{i=1}^{n}(3) + \sum_{i=1}^{n}(\frac{i}{n})])]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}(3n+\frac{i}{n}[3n + \sum_{i=1}^{n}(\frac{i}{n})])]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}(3n+\frac{i}{n}[3n + \frac{1}{n}\sum_{i=1}^{n}(i)])]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}(3n+\frac{i}{n}[3n + \frac{1}{n}\frac{n(n+1)}{2}])]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}(3n+\frac{i}{n}[3n + \frac{(n+1)}{2}])]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}(3n+3i + \frac{i(n+1)}{2n})]$

$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + 3i+\frac{3i^{2}}{n} + \frac{i^{2}(n+1)}{2n^{2}}]$

$R_{n}=lim_{n\to\infty}[1 + \frac{3i}{n}+\frac{3i^{2}}{n^{2}} + \frac{i^{2}(n+1)}{2n^{3}}]$

When considering responses, please consider that I am not a trained mathematician. I just like Mathematics.

You approach gets off track when you factor $\frac{i}{n}$ outside the sum: this is not possible, as the index $i$ is "bound" to the sum and is not defined outside. This is clear at the very end, when you have an expression still containing $i$'s -- while $i$ is not part of the limit and not defined at that point.
Instead, consider, after the second line of [4], using the (known) expressions for $\sum_{i=1}^n i$,$\sum_{i=1}^n i^2$,$\sum_{i=1}^n i^3$, as found e.g. in this other question. This will directly lead to the right expressions for the sums.
I'm not a mathematician either, but (correct me if I'm wrong) you cannot take out the $i/n$ between your 4th and 5th step as it is a sigma notation.
$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \sum_{i=1}^{n}(\frac{3i}{n}+\frac{3i^{2}}{n^{2}}+\frac{i^{3}}{n^{3}})]$
$R_{n}=lim_{n\to\infty}\frac{1}{n}[n + \frac{i}{n}\sum_{i=1}^{n}(3+\frac{3i}{n}+\frac{i^{2}}{n^{2}})]$