To find the partition of the interval \([a,b]=[-1,5]\text{,}\) we compute the partition increment size

\begin{equation*}
\Delta x = \frac{5--1}{n} = \frac{6}{n}.
\end{equation*}

The partition points are defined using an arithmetic sequence

\begin{equation*}
x_k = -1 + k \cdot \frac{6}{n} = -1 + \frac{6k}{n}.
\end{equation*}

The partition defines the \(k\)th subinterval \([x_{k-1}, x_k]\) such that the right-hand rule will evaluate the integrand \(f(x)=5-2x\) at the point \(x_k\text{,}\)

\begin{align*}
f(x_k) &= 5-2x_k = 5-2(-1+\frac{6k}{n}) \\
&= 5+2-\frac{12k}{n} = 7 - \frac{12k}{n}
\end{align*}

The Riemann sum is equal to the sum of increments computed as the integrand function (rate) times the partition increment width. That is, if we use the function \(\mathcal{R}_f(x)\) as the simple function using the right-hand end points of the intervals, then the Riemann sum is

\begin{equation*}
\int_{-1}^{5} \mathcal{R}_f(x) dx = \sum_{k=1}^{n} f(x_k) \cdot \Delta x.
\end{equation*}

Using the value we found above and \(\Delta x = \frac{6}{n}\text{,}\) this gives

\begin{align*}
\int_{-1}^{5} \mathcal{R}_f(x) dx &= \sum_{k=1}^{n} (7-\frac{12k}{n}) \cdot \frac{6}{n} \\
& = \sum_{k=1}^{n} (\frac{42}{n} - \frac{72k}{n^2}) \\
& \overset{\hbox{Linearity}}{=} \frac{1}{n} \sum_{k=1}^{n} 42 - \frac{72}{n^2} \sum_{k=1}^{n} k \\
& = \frac{1}{n} \cdot (42n) - \frac{72}{n^2} \cdot \frac{n(n+1)}{2} \\
& = 42 - \frac{36(n+1)}{n}
\end{align*}

This final formula is the value of the Riemann sum using the right-hand rule.

The limit of the Riemann sum is the value of the actual definite integral of interest. That is, for this problem, we have

\begin{align*}
\int_{-1}^{5} (5-2x) dx &= \lim_{n \to \infty} [42 - \frac{36(n+1)}{n}] \\
&= \lim_{n \to \infty} [42 - 36 \cdot \frac{n(1+\frac{1}{n})}{n}] \\
&= 42 - 36 \cdot 1 = 6.
\end{align*}

Because the graph \(y=f(x)\) (shown below) is linear, we can compute the corresponding signed area using the area of triangles and compare our calculation. The graph crosses the axis when \(f(x)=0\) which occurs at \(x=2.5\text{.}\) So we split the definite integral into two pieces,

\begin{equation*}
\int_{-1}^{5} f(x) dx = \int_{-1}^{2.5} f(x) dx + \int_{2.5}^{5} f(x) dx.
\end{equation*}

The first region on interval \([-1,2.5]\) is a triangle above the axis with height 7 and width 3.5 so that the area of the region is \(\frac{1}{2}(7)(3.5) = 12.25\text{.}\) The second region on interval \([2.5, 5]\) is a triangle below the axis with height 5 (since \(f(5)=-5\)) and base width \(5-2.5=2.5\text{.}\) The area of the second triangle is \(\frac{1}{2}(5)(2.5) = 6.25\) but corresponds to a signed area of \(-6.25\) (because below the axis). So

\begin{equation*}
\int_{-1}^{5} f(x) dx = 12.25 + -6.25 = 6.
\end{equation*}

Thus, the limit of the Riemann sum exactly agrees with the geometric calculation of total signed area.