Find the area under the graph of $f$ over the interval $[-1,4]$.
\[
f(x)=\left\{\begin{array}{ll}
x^{2}+2 & x \leq 2 \\
3 x & x> 2
\end{array}\right.
\]

Step 1 ：The function is defined differently for \(x \leq 2\) and \(x > 2\), so we will need to split the integral into two parts: one from \(-1\) to \(2\), and one from \(2\) to \(4\).

Step 2 ：For \(x \leq 2\), the function is \(f(x) = x^{2} + 2\). The area under this part of the graph from \(-1\) to \(2\) can be found by integrating this function over that interval, which gives us an area of \(9\).

Step 3 ：For \(x > 2\), the function is \(f(x) = 3x\). The area under this part of the graph from \(2\) to \(4\) can be found by integrating this function over that interval, which gives us an area of \(18\).

Step 4 ：The total area under the graph of the function over the interval \([-1,4]\) is the sum of these two areas, which is \(9 + 18 = 27\).

Step 5 ：Final Answer: The area under the graph of \(f\) over the interval \([-1,4]\) is \(\boxed{27}\).