Answer
ٍ$$
f(x)=\frac{2 x}{x^{2}+1}, \quad 1 \leqslant x \leqslant 3
$$
According to Definition 2, an expression for the area under the graph of $f$ as a limit is:
$$
\begin{aligned}
A &=\lim _{n \rightarrow \infty} R_{n} \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{2(1+2 i / n)}{(1+2 i / n)^{2}+1} \cdot \frac{2}{n}
\end{aligned}
$$
Work Step by Step
ٍ$$
f(x)=\frac{2 x}{x^{2}+1}, \quad 1 \leqslant x \leqslant 3
$$
the width of a subinterval is
$$
\Delta x=(3-1) / n=2 / n
$$
and
$$
x_{i}=1+i \Delta x=1+2 i / n
$$
The sum of the areas of the approximating rectangles is
$$
R_{n}=f\left(x_{1}\right) \Delta x+f\left(x_{2}\right) \Delta x+\cdots+f\left(x_{n}\right) \Delta x
$$
According to Definition 2, the area is
$$
\begin{aligned}
A &=\lim _{n \rightarrow \infty} R_{n} \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{2(1+2 i / n)}{(1+2 i / n)^{2}+1} \cdot \frac{2}{n}
\end{aligned}
$$
