Use the limit process to find the area under the graph of $f(x)=x^{2}+3 x$ on the interval $[0,4]$.

Step 1 ：The area under the curve of a function can be found using the limit process to sum up the areas of rectangles under the curve. This is essentially the definition of the definite integral. The area under the curve of \(f(x) = x^2 + 3x\) from \(x=0\) to \(x=4\) is given by the limit of the sum of the areas of rectangles under the curve as the width of the rectangles approaches zero. This is represented mathematically as: \[\int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x\]

Step 2 ：In this case, \(a = 0\), \(b = 4\), and \(f(x) = x^2 + 3x\). So, we have: \[\Delta x = \frac{4-0}{n} = \frac{4}{n}\] and \[x_i^* = a + i \Delta x = 0 + i \frac{4}{n} = \frac{4i}{n}\]

Step 3 ：Substituting \(x_i^*\) into \(f(x)\), we get: \[f(x_i^*) = \left(\frac{4i}{n}\right)^2 + 3\left(\frac{4i}{n}\right) = \frac{16i^2}{n^2} + \frac{12i}{n}\]

Step 4 ：So, the area under the curve is: \[\int_{0}^{4} (x^2 + 3x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} \left(\frac{16i^2}{n^2} + \frac{12i}{n}\right) \frac{4}{n}\]

Step 5 ：This is a sum of terms of the form \(ai^2 + bi\), which can be simplified using the formulas for the sum of the first \(n\) integers and the sum of the squares of the first \(n\) integers: \[\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\] and \[\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\]

Step 6 ：Using these formulas, we can simplify the sum in the limit: \[\lim_{n \to \infty} \left(\frac{16}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{12}{n} \cdot \frac{n(n+1)}{2}\right) \frac{4}{n}\]

Step 7 ：Simplifying this expression gives: \[\lim_{n \to \infty} \left(\frac{64n(n+1)(2n+1)}{6n^3} + \frac{24n(n+1)}{2n^2}\right)\]

Step 8 ：As \(n\) approaches infinity, the terms with lower powers of \(n\) in the numerator become negligible, so the limit is: \[\frac{64}{3} + 12 = \frac{100}{3}\]

Step 9 ：So, the area under the curve of \(f(x) = x^2 + 3x\) from \(x=0\) to \(x=4\) is \(\boxed{\frac{100}{3}}\) square units.