convert riemann sum $\lim_{n\to\infty}\sum_{i=1}^{n} \frac{15 \cdot \frac{3 i}{n} - 24}{n}$ to integral notation

Question

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 = $\frac {b-a}{n}$

Answer 1

I would say that the good answer is $a=-8$, $b=-5$ and $f(x)=5x+32$. Because, if you take the subdivision $(x_i)_{0\leq i\leq n}$ of $[-8,-5]$ given by :

$$x_i=-8+\frac{3i}{n} $$

you get that the associated riemann sum is :

$$\frac{3}{n}\sum_{i=1}^nf(x_i)=\frac{3}{n}\sum_{i=1}^n(5x_i+32)=\frac{3}{n}\sum_{i=1}^n(5(-8+\frac{3i}{n})+32)=...$$

$$...=\frac{3}{n}\sum_{i=1}^n(5\times \frac{3i}{n}-40+32)=\frac{3}{n}\sum_{i=1}^n(5\times \frac{3i}{n}-8) $$

Answer 2

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

$\hspace{.3 in}\displaystyle \lim_{n\to\infty}\sum_{i=1}^n \left(5\cdot\frac{3i}{n}-8\right)\frac{3}{n}=\int_0^3(5x-8)\;dx$