Answer
$\lim\limits_{n \to \infty}\sum_{i=1}^{n}[(2+\frac{3i}{n})^2+\frac{n}{2n+3i}~]~\cdot \frac{3}{n}$
Work Step by Step
$\Delta x = \frac{b-a}{n} = \frac{5-2}{n} = \frac{3}{n}$
$x_i^* = 2+\frac{3i}{n}$
We can express the integral as a limit of Riemann sums:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i^*)\Delta x$
$\int_{1}^{3} (x^2+\frac{1}{x}) ~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}[(2+\frac{3i}{n})^2+\frac{1}{2+\frac{3i}{n}}~]~\cdot \frac{3}{n}$
$\int_{1}^{3} (x^2+\frac{1}{x}) ~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}[(2+\frac{3i}{n})^2+\frac{n}{2n+3i}~]~\cdot \frac{3}{n}$
