We need to evaluate the definite integral and subtract a constant. The expression is: $A(R) = \int_{-4}^{3} [3 - x - (x^2 - 9)] dx - \frac{9\pi}{2}$.

1. Problem Description

We need to evaluate the definite integral and subtract a constant. The expression is:

$A(R) = \int_{-4}^{3} [3 - x - (x^2 - 9)] dx - \frac{9\pi}{2}$.

2. Solution Steps

First, simplify the integrand:

$3 - x - (x^2 - 9) = 3 - x - x^2 + 9 = 12 - x - x^2$.

Now, integrate the simplified integrand with respect to $x$:

$\int (12 - x - x^2) dx = 12x - \frac{x^2}{2} - \frac{x^3}{3} + C$.

Next, evaluate the definite integral from -4 to 3:

$\int_{-4}^{3} (12 - x - x^2) dx = \left[12x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-4}^{3}$

$= \left(12(3) - \frac{(3)^2}{2} - \frac{(3)^3}{3}\right) - \left(12(-4) - \frac{(-4)^2}{2} - \frac{(-4)^3}{3}\right)$

$= \left(36 - \frac{9}{2} - \frac{27}{3}\right) - \left(-48 - \frac{16}{2} - \frac{-64}{3}\right)$

$= \left(36 - \frac{9}{2} - 9\right) - \left(-48 - 8 + \frac{64}{3}\right)$

$= \left(27 - \frac{9}{2}\right) - \left(-56 + \frac{64}{3}\right)$

$= 27 - \frac{9}{2} + 56 - \frac{64}{3}$

$= 83 - \frac{9}{2} - \frac{64}{3}$

$= 83 - \frac{27}{6} - \frac{128}{6}$

$= 83 - \frac{155}{6}$

$= \frac{498}{6} - \frac{155}{6}$

$= \frac{343}{6}$.

Finally, subtract $\frac{9\pi}{2}$ from the result:

$A(R) = \frac{343}{6} - \frac{9\pi}{2}$

$= \frac{343}{6} - \frac{27\pi}{6}$

$= \frac{343 - 27\pi}{6}$.

3. Final Answer

$\frac{343 - 27\pi}{6}$