Hello I'm currently studying for my Calculus 2 exam and i have a pretty good grasp up until now. I'm having problems finding a good explanation on how to solve limits with Riemann sums. Can someone please explain for this particular problem. All other problems are similar to this one so if i know how to solve it i can solve all of them.
The problem in question is ,
$\lim_{n\to+\infty} \sum_{k=1}^n \frac{k^2+n^2}{n^3+1} $
To evaluate $$\lim_{n\to\infty}\sum_{k=1}^n{\frac{k^2+n^2}{n^3+1}}$$ by the method of Riemann sums, note that we need to express the sum as an average, so we will divide by $n$, which means we will want $\Delta x$ to equal $1/n$. Unfortunately, we have $n^3+1$ in the denominator, but we can replace it by $n^3,$ since $$\lim_{n\to\infty}\frac{n^3}{n^3+1}=1$$
So now, we can write $$ \lim_{n\to\infty}\sum_{k=1}^n{\frac{k^2+n^2}{n^3+1}}= \lim_{n\to\infty}\sum_{k=1}^n{\frac{k^2+n^2}{n^3}}= \lim_{\Delta x\to 0+}\sum_{k=1}^{n}{\Delta x}(1+(k \Delta x)^2)=\int_0^1{(1+x^2)dx} $$
EDIT Following up on the OP's comment, sometimes you have to be flexible. Suppose that we had $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^n{\frac{k^2+n^2}{n^2+1}}$$
Now we have the divisor given to us, but we're dividing by $n$ and there are $n+1$ terms. We can pull out the $k=0$ term however, to get $$\lim_{n\to\infty}\frac{n}{n^2+1}+ \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n{\frac{k^2+n^2}{n^2+1}}= \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n{\frac{k^2+n^2}{n^2}}= \int_0^1{(1+x^2)dx} $$