This is 1 of the 3 remaining questions I have left due for my webwork which is due tonight. If someone can please show me how to do this I can do the following two which are based on the same concept. People have told me how to partially do it but I just can't really tie it to this question. Thanks in advance.
Right Riemann sums are of the form: $\sum_{i=1}^nf(a+i\Delta x)*\Delta x$, where $\Delta x = \frac {b-a}n$, $b$ is your right endpoint, $a$ is your left endpoint, and $n$ is the number of rectangles you have. To fit this to your problem, look for what specifically is the $\Delta x$. The $\Delta x$ should be the same in every term.
Example: I am using a different function set and numbers to avoid any cheating.
$$\sin\left(0+\frac2n\right)*\left(\frac2n\right)+\sin\left(0+\frac4n\right)*\left(\frac2n\right)+\sin\left(0+\frac6n\right)*\left(\frac2n\right)+...+\sin\left(0+\frac{2n}n\right)*\left(\frac2n\right)$$
In this case, $a=0$, $\Delta x=\frac2n=\frac{b-0}n$, and $b=2$. You can find out what $\Delta x$ is by looking at the term that remains constant throughout the sum and by the difference between the arguments of the function, i.e. $\frac4n-\frac2n=\frac2n$. You can then use $\Delta x$ to find $b$ by adding it to $a$.