How to express a Riemann sum as a definite integral For this function, I know that the a value is 1 and that the $\Delta x$ is $\frac{1}{n}$. I'm not sure how to go from there in order to express this as a definite integral. Can anyone please help me out?
$$\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n\frac{3}{1+(\frac{i}{n})^2}$$
 A: One way to express the Riemann sum as a definite integral is
$$\int_a^bf(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(a+i\,\Delta x)\,\Delta x$$ where $$\Delta x =\frac{b-a}n$$
Taking $$J=\lim_{n\to\infty}\frac1n\sum_{i=1}^n\frac3{1+\left(\frac in\right)^2}$$ and making the substitution $g(u)=1/(1+u^2)$ with some rearrangement gives
$$J=3\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^ng\!\left(\frac in\right)$$
Take $\Delta x=1/n$, and at this point you can probably see that $a=0$ and $b=1$:
$$\begin{align}
J&=3\lim_{n\to\infty}\frac{1-0}{n}\sum_{i=1}^n g\!\left( 0+i\frac{1-0}{n}\right) \\
&=3\lim_{n\to\infty}\Delta x\sum_{i=1}^n g(a+i\,\Delta x)
\end{align}$$
and thusly you get $$J=3\int_0^1 \frac{dx}{1+x^2}$$
This graph clearly suggests that this is in fact a correct reformulation of $J$.
A: You want to see the $n^{th}$ term in that sequence that you are taking the limit from as the $n^{th}$ Riemann sum. The $\frac{1}{n}$ tells you that you divide your interval in $n$ equal parts. Then, by definition of Riemann sums, each term in the sum must be the value of your function either at the beginning or at the end of each part of the interval in which you divided it. The thing that varies term by term is what is inside the square, so this already suggests that this is the $x$. And now it reminds to compute the limts of integration. In the last term of the sum, you get $\frac{n}{n}=1$, so take this as the upper limit (hence we have a right Riemann sum). This makes the value in the first term, $\frac{1}{n}$, the value of $x$ at the right end of our first segment of interval (of length $\frac{1}{n}$). This shows that the left integration limit must be $0$. So we have:
$$\int_{0}^{1}\frac{3}{1+x^{2}}= \lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^{n}\frac{3}{1+(\frac{i}{n})^{2}}$$
