Find the area between the curves.
\[
x=-5, x=3, y=3 x, y=x^{2}-4
\]
The area between the curves is (Simplify your answer.)

Step 1 ：First, find the points of intersection between the curves \(y=3x\) and \(y=x^2-4\) by setting \(3x = x^2 - 4\), which gives \(x^2 - 3x - 4 = 0\).

Step 2 ：Factor the equation to get \((x - 4)(x + 1) = 0\), so \(x = 4\) and \(x = -1\) are the points of intersection.

Step 3 ：However, we are only interested in the area between \(x = -5\) and \(x = 3\), so we discard \(x = 4\) as it is outside this range.

Step 4 ：The area between the curves is given by the integral of the absolute difference of the two functions from the lower limit to the upper limit.

Step 5 ：So, the area \(A\) is given by \(A = \int_{-5}^{-1} |3x - (x^2 - 4)| dx + \int_{-1}^{3} |(x^2 - 4) - 3x| dx\).

Step 6 ：Solving the integrals, we get \(A = \int_{-5}^{-1} (3x - x^2 + 4) dx + \int_{-1}^{3} (x^2 - 3x + 4) dx\).

Step 7 ：This simplifies to \(A = [1.5x^2 - (1/3)x^3 + 4x]_{-5}^{-1} + [(1/3)x^3 - 1.5x^2 + 4x]_{-1}^{3}\).

Step 8 ：Evaluating the integrals gives \(A = [1.5*(-1)^2 - (1/3)*(-1)^3 + 4*(-1)] - [1.5*(-5)^2 - (1/3)*(-5)^3 + 4*(-5)] + [(1/3)*3^3 - 1.5*3^2 + 4*3] - [(1/3)*(-1)^3 - 1.5*(-1)^2 + 4*(-1)]\).

Step 9 ：Simplifying further gives \(A = [1.5 - 1/3 - 4] - [37.5 + 125/3 - 20] + [9 - 13.5 + 12] - [-1/3 - 1.5 + 4]\).

Step 10 ：Finally, adding these values gives \(A = -2.833 + 92.833 + 7.5 + 2.833\), which simplifies to \(\boxed{100.333}\) square units.