I've taught definite integrals for high school twice, but everytime I run into this particular Riemann Sum to Definite Integral question that I can't seem to solve.
The question is:
If $n$ is a positive integer, then $\lim\limits_{n\to \infty} \frac{4}{n} \bigg[ \bigg(2+ \frac{4}{n}\bigg)+ \bigg(2+ \frac{5}{n}\bigg) + \bigg(2+ \frac{6}{n}\bigg) + \ldots \bigg(2+ \frac{n}{n}\bigg) \bigg]$ can be expressed as what definite integral?
I have attempted a solution by letting $\Delta x= \frac{4}{n}=(b-a)/n$ and based on the function inside, it seems like $f(x)=x$. Trying to find the bounds of integration is where my solution is dependent on my assumption since I usually assume $a=2$ based on how the sum is written. This does not seem right.
After attempting it once again, I ended up with something like $\lim\limits_{n\to \infty} \sum\limits_{k=1}^n \bigg[ \bigg(2+ \frac{k(1+\frac{3}{k})}{n}\bigg) \bigg] \frac{4}{n}$ Thanks!
$\int_0^{1}4(2+x)dx=\lim \frac 1 n \sum\limits_{i=1}^{n} 4(2+\frac i n)$. This is same as the given limit since the limit of $\frac 1 n 4(2+\frac i n)$ is $0$ for $i\leq 3$. So the answer is $10$.