# Evaluate Integral $\int_{1}^{3}(3x + 2)dx$ as a Riemann Sum

I am having difficulties working out this Riemann Sum.

$$\Delta x = \frac{2}{n},~~ x_i = 1 + \frac{2i}{n},~~ i = \frac{n(n+1)}{2}$$
where $$\int_{1}^{3}(3x + 2)dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i) \Delta x$$ $$~$$ $$~$$ $$= \lim_{n\to\infty} \sum_{i=1}^{n} \bigg( \frac{3+6i}{n} + 2 \bigg) \frac{2}{n} = \lim_{n\to\infty} \sum_{i=1}^{n}\bigg(3\big(1 + \frac{2i}{n}\big) + 2\bigg) \frac{2}{n}$$

This is how my teacher answered the problem.

$$= \lim_{n\to\infty} \frac{2}{n}^{***} \sum_{i=1}^{n} 3 + \frac{6i}{n} + 2 = \lim_{n\to\infty} \frac{2}{n} \sum_{i=1}^{n} 5 + \frac{6i}{n}$$

$$= \lim_{n\to\infty} \frac{2}{n} 5n^{****} + \frac{2}{n} \sum_{i=1}^{n} \frac{6i}{n} = \lim_{n\to\infty} 10 + \frac{12}{n^2}^{*****} \sum_{i=1}^{n}i = 10 + \lim_{n\to\infty} \frac{12}{n^2} \frac{n(n+1)}{2}$$ $$= 10 + \lim_{n\to\infty} 6(1+\frac{1}{n}) = 10 + 6 = 16$$

I think all of my questions stem from a lack of understanding from my first question:

*** What property lets you move the $$\frac{2}{n}$$ before the sum? Is this just the associative property at work here?
**** This is where I get lost, I don't get what step is happening here and why.
***** why is $$\frac{2}{n} and \frac{6i}{n}$$ allowed to combine without distributing the $$\frac{6i}{n}$$ over to the 10 as well?

• What do you need $i = \frac{n(n+1)}{2}$ for? – Michael Rybkin Feb 5 at 20:28
• i is the sum of the integers telescoping sum, I believe it is used so that there is only one variable $n$ – Evan Kim Feb 5 at 20:32

You can pull $$n$$ in and out like that because for all intents and purposes it's a constant with respect to the summation symbol. $$i$$ is what's changing. You cannot, however, pull $$n$$ outside the limit symbol.
\begin{align} \lim_{n\to\infty} \sum_{i=1}^{n}\bigg[3\bigg(1 + \frac{2i}{n}\bigg) + 2\bigg] \frac{2}{n} &=\lim_{n\to\infty} \sum_{i=1}^{n}\bigg(3 + \frac{6i}{n} + 2\bigg) \frac{2}{n}\\ &=2\lim_{n\to\infty} \sum_{i=1}^{n}\bigg(5 + \frac{6i}{n}\bigg) \frac{1}{n}\\ &=2\lim_{n\to\infty} \sum_{i=1}^{n}\bigg(\frac{5}{n} + \frac{6i}{n^2}\bigg)\\ &=2\lim_{n\to\infty}\bigg(\frac{5}{n}\sum_{i=1}^{n}1 + \frac{6}{n^2}\sum_{i=1}^{n}i\bigg)\\ &=2\lim_{n\to\infty}\bigg(\frac{5}{n}\cdot n + \frac{6}{n^2}\cdot\frac{n(n+1)}{2}\bigg)\\ &=2\lim_{n\to\infty}\bigg(5 + \frac{3+3n}{n}\bigg)\\ &=2\lim_{n\to\infty}\bigg(5 + \frac{3}{n}+3\bigg)\\ &=2\lim_{n\to\infty}\bigg(8 + \frac{3}{n}\bigg)\\ &=2\cdot(8+0)\\ &=2\cdot 8\\ &=16 \end{align}
• Thanks, that makes more sense. Towards the middle, I see that you have $\frac{5}{n} \dot n$. The $\sum_{i=1}^{n}$ converted to $n$? – Evan Kim Feb 5 at 21:18
• Well, what does $\sum\limits_{i=1}^{n}1$ mean? $\sum\limits_{i=1}^{n}1=1+1+1+...+1$. It's 1 taken $n$ times. That's going to equal $n$. – Michael Rybkin Feb 5 at 21:21
Perhaps this will help: think about a fixed value of $$n$$, and examine $$\sum_{i=1}^{n}\bigg(3\big(1 + \frac{2i}{n}\big) + 2\bigg) \frac{2}{n} = \sum_{i=1}^n \frac{2}{n}\cdot 5 + \sum_{i=1}^n \frac{2}{n}\cdot\frac{6i}{n} = \sum_{i=1}^n \frac{10}{n} + \sum_{i=1}^n \frac{12}{n^2}\cdot i.$$ Since we are thinking about a fixed $$n$$, terms involving only $$n$$ are essentially constants and can be pulled out of the summation, just like a factor of $$13$$ (for example) could be. So we get $$\frac{10}{n}\cdot\sum_{i=1}^n 1 + \frac{12}{n^2}\cdot\sum_{i=1}^n i.$$