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We need to find the area between $y = 4 - x^2$ and the $x$-axis.

Step 1: Find where the curve crosses the x-axis. Set $y = 0$: $4 - x^2 = 0$, so $x^2 = 4 \implies x = \pm 2$. So the limits of integration are $-2$ and $2$.

Step 2: Set up the integral. The area is the integral of the function above the $x$-axis: $A = \int_{-2}^{2} (4 - x^2)\,dx$

Step 3: Integrate. Using the power rule, $\int (4 - x^2)\,dx = 4x - \frac{x^3}{3}$. Evaluate at the limits: $A = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \frac{16}{3} - \left(-\frac{16}{3}\right)$

Final answer: $A = \frac{32}{3}$ square units

No problem! In the last step we had $\frac{16}{3} - \left(-\frac{16}{3}\right)$. Subtracting a negative is the same as adding, so it becomes $\frac{16}{3} + \frac{16}{3}$.

Both have the same denominator, so add the tops: $\frac{16 + 16}{3} = \frac{32}{3} \approx 10.67$ square units.