Write the Riemann sum for $f(x)=\sqrt{x}$ on the interval $[5,30]$, using 20 subintervals of equal width, and taking $x_{i}^{*}$ to be the left endpoint of the $i^{\text {th }}$ subinterval.
First, compute $\Delta x$ and give a formula for $x_{i}^{*}$.
\[
\begin{array}{l}
\Delta x=\frac{5}{4} \quad \\
x_{i}^{*}=\square
\end{array}
\]

Step 1 ：First, we calculate the width of each subinterval, denoted as \(\Delta x\). This is given by the formula \(\Delta x = \frac{b - a}{n}\), where \(a\) and \(b\) are the endpoints of the interval and \(n\) is the number of subintervals. In this case, \(a = 5\), \(b = 30\), and \(n = 20\). So, \(\Delta x = \frac{30 - 5}{20} = \frac{25}{20} = 1.25\).

Step 2 ：Next, we find a formula for \(x_{i}^{*}\), which is the left endpoint of the \(i^{th}\) subinterval. Since we are starting at \(a = 5\) and each subinterval has width \(\Delta x = 1.25\), the left endpoint of the \(i^{th}\) subinterval is given by \(x_{i}^{*} = a + (i - 1)\Delta x = 5 + (i - 1)1.25\).

Step 3 ：Finally, the Riemann sum is given by the sum of the function values at these left endpoints, multiplied by the width of the subintervals. In sigma notation, this is \(\sum_{i=1}^{20} f(x_{i}^{*}) \Delta x\), or in this case, \(\sum_{i=1}^{20} \sqrt{5 + (i - 1)1.25} \cdot 1.25\).

Step 4 ：So, the Riemann sum for \(f(x) = \sqrt{x}\) on the interval \([5,30]\), using 20 subintervals of equal width, and taking \(x_{i}^{*}\) to be the left endpoint of the \(i^{th}\) subinterval, is \(\sum_{i=1}^{20} \sqrt{5 + (i - 1)1.25} \cdot 1.25\).