For the function given below, find a formula for the Riemann sum obtained by dividing the interval $[a, b]$ into $n$ equal subintervals and using the right-hand endpoint for each $c_{k}$. Then take a limit of this sum as $n \rightarrow \infty$ to calculate the area under the curve over $[a, b]$. $f(x)=3 x$ over the interval $[2,6]$ Find a formula for the Riemann sum. \[ S_{n}= \]

Step 1 ：Given the function \(f(x) = 3x\) over the interval \([2, 6]\), we are asked to find a formula for the Riemann sum obtained by dividing the interval into \(n\) equal subintervals and using the right-hand endpoint for each \(c_{k}\).

Step 2 ：The formula for the Riemann sum is given by \(S_{n} = \sum_{k=1}^{n} f(a + k \cdot \Delta x) \cdot \Delta x\), where \(\Delta x = \frac{b - a}{n}\) is the width of each subinterval, and \(f(a + k \cdot \Delta x)\) is the height of the \(k\)-th rectangle.

Step 3 ：In this case, \(a = 2\), \(b = 6\), and \(f(x) = 3x\). So, \(\Delta x = \frac{6 - 2}{n} = \frac{4}{n}\), and \(f(a + k \cdot \Delta x) = 3(a + k \cdot \Delta x) = 3(2 + k \cdot \frac{4}{n}) = \frac{12k}{n} + 6\).

Step 4 ：Substituting these values into the formula for the Riemann sum, we get \(S_{n} = \sum_{k=1}^{n} \left(\frac{12k}{n} + 6\right) \cdot \frac{4}{n} = 24 + 48\left(\frac{n^{2}}{2} + \frac{n}{2}\right)/n^{2}\).

Step 5 ：Now, we need to take the limit of this sum as \(n \rightarrow \infty\) to calculate the area under the curve over \([2, 6]\).

Step 6 ：As \(n \rightarrow \infty\), the term \(\frac{n^{2}}{2} + \frac{n}{2}\) in the Riemann sum becomes very large, and the term \(24 + 48\left(\frac{n^{2}}{2} + \frac{n}{2}\right)/n^{2}\) approaches 48.

Step 7 ：Final Answer: The area under the curve \(f(x) = 3x\) over the interval \([2, 6]\) is \(\boxed{48}\).