I am a bit struggling with transforming a Riemann's sum into its definite integral form.
I am given the following equation:
$$ \sum_{i=1}^n\left(\left(3-\frac{i}{n}\right)^3\right) \frac{1}{n} $$
Now, so far I have understood that to transform a Riemann sum we need to use the following rule:
$$ \lim _{n \rightarrow \infty} \sum_{i=1}^n f(a+i \Delta x) \Delta x=\int_a^b f(x) d x $$ From this, we can equate matching parts of both equations:
- $a$ from $a+i \Delta x = 3-\frac{i}{n}$, which means that $a=3$;
- $b$ from $\Delta x=\frac{1}{n}$, and since $\Delta x$ is also equal to $\frac{b-a}{n}$, $b-a=1$, therefore $b=4$.
- $f(x)$, which is where a have the problem, (assuming I've done the rest correctly).
Since I don't have an explicit function here like say $\lim _{n \rightarrow \infty} \sum_{i=1}^n ln(2+\frac{5i}{n})$, where we can immediately see that $f(x)$ is $ln(x)$, I thought that in my case the function is probably the power inside the parenthesis, or $x^3$. Is that correct, or am I wrong here?
Using this, I got $\int_3^4 x^3 d x$, which is $\frac{175}{4}=43.75$, which is different from the converging value of the Riemann sum when $n=1000$, which is around $16.24$, so I must be wrong somewhere.