Answer
\begin{align}
\lim\limits_{max \ \varDelta x_{k} \to 0} \sum_{k=1}^{n} (3x_{k}^{*} +1) \varDelta x_{k} = \frac{5}{2}
\end{align}
Work Step by Step
The given Riemann sum is
\begin{align}
\lim\limits_{max \ \varDelta x_{k} \to 0} \sum_{k=1}^{n} (3x_{k}^{*} +1) \varDelta x_{k}
\end{align}
The definite integral form of the given Riemann sum is
\begin{alignat}{1}
\int_{0}^{1} (3x+1) \ dx = \big[3\frac{x^{2}}{2} + x \big]_{0}^{1} = \frac{5}{2}
\end{alignat}
