I'm doing a question where I need to prove that a function, $x^2 + 2x$ is Riemann integrable on the interval $[-3,3]$.
The question also gives me the interval $2n$, first $n$ intervals in $[-3,-1]$ with the width of $2/n$ and second $n$ intervals in $[-1,3]$ with the width of $4/n$.
I have already been given the suggested workings but I cannot seem to understand.
The workings given to me for lower Riemann are as follows:
$$ \sum_{i=1}^n f\left(-1-\frac{2i}{n}\right)\frac{2}{n} + \sum_{i=1}^n f\left(-1-\frac{4i}{n}\right)\frac{4}{n} $$
Which ends at $18-\frac{36}{n}+\frac{12}{n^2}$ which equals to 18 when $n$ approaches infinity.
I can understand the intervals $\frac{2}{n}$ and $\frac{4}{n}$. But I cannot understand the brackets with $(-1-\frac{2i}{n})$ and $(-1-\frac{4i}{n})$. Why is $-1$ the start and not $-3$?
hint
If the function $ f $ is Riemann integrable at $ [a,b] $ then
$$\lim_{n\to+\infty}\frac{b-a}{n}\sum_{i=1}^nf(a+i\frac{b-a}{n})=\int_a^bf(x)dx$$
With $ (a,b)=(-3,-1) $, the sum becomes $$\frac 2n\sum_{i=1}^nf(-3+i\frac 2n)$$
and with $ (a,b)=(-1,3) $, it gives $$\frac 4n\sum_{i=1}^nf(-1+i\frac 4n)$$