Find the area of the region enclosed by $f(x)$ and the $x$-axis for the given function over the specified interval.
\[
f(x)=\left\{\begin{array}{ll}
x^{2}+3 x+4 & x< 3 \\
4 x+10 & x \geq 3
\end{array} \text { on }[-6,6]\right.
\]

Step 1 ：The function is defined differently for \(x<3\) and \(x\geq3\), so we need to split the integral into two parts: one from \(-6\) to \(3\) and the other from \(3\) to \(6\).

Step 2 ：For \(x<3\), the function is \(f(x) = x^{2} + 3x + 4\). We integrate this from \(-6\) to \(3\) to find the area under the curve and above the x-axis.

Step 3 ：The result of this integration is \(\frac{153}{2}\).

Step 4 ：For \(x\geq3\), the function is \(f(x) = 4x + 10\). We integrate this from \(3\) to \(6\) to find the area under the curve and above the x-axis.

Step 5 ：The result of this integration is \(84\).

Step 6 ：We add these two areas together to find the total area enclosed by the function and the x-axis over the interval \([-6,6]\).

Step 7 ：The total area is \(\frac{153}{2} + 84 = \frac{321}{2}\).

Step 8 ：Final Answer: The area of the region enclosed by \(f(x)\) and the \(x\)-axis for the given function over the specified interval is \(\boxed{\frac{321}{2}}\).