Answer
$\int_1^3\sqrt{4+x^2}dx=\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{2}{n}\sqrt{\frac{5n^2+4in+4i^2}{n^2}}$
Work Step by Step
Step 1: Find $\Delta x$ and $x_i=a+i\Delta x$.
The lower bound: $a=1$
The upper bound: $b=3$
$\Delta x=\frac{b-a}{n}=\frac{3-1}{n}=\frac{2}{n}$
$x_i=1+i\cdot \frac{2}{n}$
$x_i=1+\frac{2i}{n}$
Step 2: Find $R_n$.
$R_n=\sum_{i=1}^n\sqrt{4+x_i^2}\cdot \frac{2}{n}$
$R_n=\sum_{i=1}^n\sqrt{4+(1+\frac{2i}{n})^2}\cdot \frac{2}{n}$
$R_n=\sum_{i=1}^n\sqrt{4+1+\frac{4i}{n}+\frac{4i^2}{n^2}}\cdot \frac{2}{n}$
$R_n=\sum_{i=1}^n\frac{2}{n}\sqrt{\frac{5n^2+4in+4i^2}{n^2}}$
Step 3: Write the define as a limit of Riemann sum using right endpoints.
$\int_1^3\sqrt{4+x^2}dx=\lim\limits_{n \to \infty}R_n$
$\int_1^3\sqrt{4+x^2}dx=\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{2}{n}\sqrt{\frac{5n^2+4in+4i^2}{n^2}}$