Step 1: Find where the curve meets the x-axis.
$y=x^2-4=0$ gives $x=\pm2$. Between $-2$ and $2$ the curve is below the axis; from $2$ to $3$ it is above.
Step 2: Set up the area with correct signs.
Area is always positive, so \[ A=-\int_{-2}^{2}(x^2-4)\,dx+\int_{2}^{3}(x^2-4)\,dx. \]
Step 3: Evaluate the first part.
\[ -\int_{-2}^{2}(x^2-4)\,dx=\int_{-2}^{2}(4-x^2)\,dx=\left[4x-\frac{x^3}{3}\right]_{-2}^{2}. \]
Step 4: Plug in the limits.
$\left(8-\dfrac83\right)-\left(-8+\dfrac83\right)=\dfrac{16}{3}+\dfrac{16}{3}=\dfrac{32}{3}$.
Step 5: Evaluate the second part.
\[ \int_{2}^{3}(x^2-4)\,dx=\left[\frac{x^3}{3}-4x\right]_{2}^{3}=(9-12)-\left(\frac83-8\right)=-3+\frac{16}{3}=\frac{7}{3}. \]
Step 6: Add the pieces.
\[ A=\frac{32}{3}+\frac{7}{3}=\frac{39}{3}=13. \] \[ \boxed{13} \]