Determining the infinite limit of a Riemann' sum I need to evaluate the following $f(x)=x^2 + 2x - 5$, on $[1,4]$, by using the riemans sum and limiting it to infinity. I have set up everything $\Delta x=\frac{3}{n}$.
$x_i=1+\frac{3i}{n}$, I would like to just check my workings against someone who actually finds this fairly simple, the part which i struggle with the most is the simplifying of the summation.
Pg. 1/2 working out
Pg. 2/2 working out
P.S: Apologies for untidy hand writing 
 A: My suggestion is that when you calculate an integral using Riemann sums it might be easier to find $\int_0^4 f-\int_0^1 f$ than $\int_1^4 f$ in one step -avoiding terms of the form $(h+\frac{k}{n})^c$-.
Let $R_n^{4}$  be the Riemann sum of n-terms from $0$ to $1 $ and $R_n^1$ the Riemann sum from $0$ to $1$. It should be fairly straight forward to compute each one of these separately.
$$R_n^4=\sum_{i=1}^n f\left(\frac{4}{n}\right)\left(\frac{4}{n}\right)\\=\sum_{i=1}^n\left(\frac{16}{n^2}i^2+\frac{8}{n}i-5\right)\left(\frac{4}{n}\right)\\=\left(\frac{64}{n^3}\right)\sum_{i=1}^n i^2+\frac{32}{n^2}\sum_{i=1}^n i-\frac{20}{n}\sum_{i=1}^n 1\\=\left(\frac{64}{n^3}\right)\left(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right)+\left(\frac{32}{n^2}\right)\left(\frac{n(n+1)}{2}\right)-\frac{20}{n}n\\=\frac{64}{3}+\frac{64}{2n}+\frac{64}{6n^2}+16+\frac{16}{n}-20.$$
Take $n\to+\infty$ and you get $\int_0^4=\frac{64}{3}-4$. And
$$R^1_n=\sum_{i=1}^n f\left(\frac{1}{n}\right)\left(\frac{1}{n}\right)\\=\sum_{i=1}^n \left(\frac{1}{n^2}i^2+\frac{2}{n}i-5\right)\left(\frac{1}{n}\right)\\=\frac{1}{n^3}\left(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right)+\frac{2}{n^2}\left(\frac{n(n+1)}{2}\right)-5\\=\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}+1+\frac{1}{n}-5.$$
Taking $n\to +\infty$ we get $\int_0^1=\frac{1}{3}-4$
