Divide the interval [3, 5] into [tex]n[/tex] subintervals of equal length [tex]\Delta x=\frac{5-3}n = \frac2n[/tex].
[tex][3,5] = \left[3+\dfrac0n,3+\dfrac2n\right] \cup \left[3+\dfrac2n,3+\dfrac4n\right]\cup\left[3+\dfrac4n,3+\dfrac6n\right]\cup\cdots\cup\left[3+\dfrac{2(n-1)}n, 3+\dfrac{2n}n\right][/tex]
The right endpoint of the [tex]i[/tex]-th subinterval is
[tex]r_i = 3 + \dfrac{2i}n[/tex]
where [tex]1\le i\le n[/tex].
Then the definite integral is given by the Riemann sum
[tex]\displaystyle \int_3^5 \sqrt{8+x^2} \, dx = \lim_{n\to\infty} \sum_{i=1}^n \sqrt{8+{r_i}^2} \Delta x = \boxed{\lim_{n\to\infty} \frac2n \sum_{i=1}^n \sqrt{17 + \frac{12i}n + \frac{4i^2}{n^2}}}[/tex]
