Area Between Curve And Axis

Finding the area between a downward parabola and the x-axis begins with locating the roots, the points where the curve crosses the axis and bounds the region. Solving 4 - x^2 = 0 gives x = -2 and x = 2 as the limits of integration. Since 4 - x^2 \ge 0 throughout [-2, 2], the function stays above the axis and the area equals \int_{-2}^{2}(4 - x^2),dx directly. Because the integrand is even, we exploit symmetry about the y-axis and write this as 2\int_0^2 (4 - x^2),dx = 2\left[4x - \frac{x^3}{3}\right]_0^2 = 2\left(8 - \frac{8}{3}\right) = 2\cdot\frac{16}{3} = \frac{32}{3}. Option 16/3 keeps only the half-interval result and forgets to double. Option 8 evaluates the 4x term alone, dropping the cubic piece. Option 64/3 doubles one time too many. The even-symmetry shortcut neatly halves the arithmetic for symmetric integrands. As a final plausibility check, the parabola peaks at height 4 over a base of width 4, bounding a rectangle of area 16, and the value 32/3 \approx 10.67 sensibly fills about two-thirds of that rectangle, which is exactly the fraction a parabolic cap is known to occupy of its enclosing rectangle.