Answer
$\int_{1}^{3} \frac{x}{x^2+4}~dx$
Work Step by Step
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\Sigma_{i=1}^{n} f(x)~\Delta x$
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\Sigma_{i=1}^{n} \frac{x_i^*}{(x_i^*)^2+4}~\Delta x$
In the limit, we can see that the function is $f(x) = \frac{x}{x^2+4}$
We can write the definite interval:
$\int_{1}^{3} \frac{x}{x^2+4}~dx$