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

Step 1 ：The area enclosed by a function \(f(x)\) and the \(x\)-axis over an interval \([a, b]\) can be found by integrating the absolute value of the function from \(a\) to \(b\). In this case, we have two functions to consider: \(f_1(x) = x^2 + 2x + 2\) for \(x < 2\) and \(f_2(x) = 3x + 4\) for \(x \geq 2\). We need to calculate the area under each function separately and then add them together.

Step 2 ：The area under \(f_1(x)\) from \(-6\) to \(2\) is given by \(\int_{-6}^{2} |f_1(x)| dx\) and the area under \(f_2(x)\) from \(2\) to \(4\) is given by \(\int_{2}^{4} |f_2(x)| dx\).

Step 3 ：Since both \(f_1(x)\) and \(f_2(x)\) are always positive over the given intervals, we can drop the absolute value signs in the integrals.

Step 4 ：The area under the first function \(f_1(x)\) from \(-6\) to \(2\) is \(\frac{176}{3}\).

Step 5 ：The area under the second function \(f_2(x)\) from \(2\) to \(4\) is \(26\).

Step 6 ：Adding these two areas together gives a total area of \(\frac{254}{3}\).

Step 7 ：Final Answer: The area of the region enclosed by the function \(f(x)\) and the \(x\)-axis over the interval \([-6, 4]\) is \(\boxed{\frac{254}{3}}\).