convert riemann sum $\lim_{n\to\infty}\sum_{i=1}^{n} \frac{15 \cdot \frac{3 i}{n} - 24}{n}$ to integral notation The limit 
$ \quad\quad \displaystyle \lim_{n\to\infty}\sum_{i=1}^{n} \frac{15 \cdot \frac{3 i}{n} - 24}{n} $
is the limit of a Riemann sum for a certain definite integral 
$
 \quad\quad \displaystyle \int_a^b f(x)\, dx $
What are the values of:
a = 
b = 
f(x) = 
?
I said:
a = -8
b = -5
f(x) = 5x
Why is this not correct? It checks out.
So I rewrote the riemann sum notation like this:
$\frac {3}{n} \cdot (\frac {3i}{n} \cdot 5 - 8) $
as you can see Δx is $\frac {3}{n}$
since my a is -8, my b is therefore -5 because Δx = $\frac {b-a}{n}$

Here are some alternate answers:
1) Take $a=0, b=1$; using n equal subintervals and right endpoints as sampling numbers, we get that
$\hspace{.3 in}\displaystyle \lim_{n\to\infty}\sum_{i=1}^n \left(45\cdot\frac{i}{n}-24\right)\frac{1}{n}=\int_0^1(45x-24)\;dx$
2) Take $a=0, b=3$; using n equal subintervals and right endpoints as sampling numbers, we get that
$\hspace{.3 in}\displaystyle \lim_{n\to\infty}\sum_{i=1}^n \left(5\cdot\frac{3i}{n}-8\right)\frac{3}{n}=\int_0^3(5x-8)\;dx$
