Formula for Riemann Sum is:
[tex]\frac{b-a}{n} \sum_{i=1}^n f(a + i \frac{b-a}{n})[/tex]
interval is [1,3] so a = 1, b = 3
f(x) = 3x , sub into Riemann sum
[tex]\frac{2}{n} \sum_{i=1}^n 3(1 + \frac{2i}{n})[/tex]
Continue by simplifying using properties of summations.
[tex]= \frac{2}{n}\sum_{i=1}^n 3 + \frac{2}{n}\sum_{i=1}^n \frac{6i}{n} \\ \\ = \frac{6}{n}\sum_{i=1}^n 1 + \frac{12}{n^2}\sum_{i=1}^n i \\ \\ =\frac{6}{n} (n) + \frac{12}{n^2}(\frac{n(n+1)}{2}) \\ \\ =6+\frac{6}{n}(n+1) \\ \\ =12 + \frac{6}{n} [/tex]
Now you have an expression for the summation in terms of 'n'.
Next, take the limit as n-> infinity.
The limit of [tex]\frac{6}{n}[/tex] goes to 0, therefore the limit of the summation is 12.
The area under the curve from [1,3] is equal to limit of summation which is 12.
