Question

Find the area of the region bounded by the curve \( y = x^2 \) and the line \( y = 4 \).

Hint:

For area between curves, integrate the difference over intersection points; use symmetry for even functions.

Answer 1

The curve \( y = x^2 \) intersects \( y = 4 \) at \( x^2 = 4 \Rightarrow x = \pm 2 \). Area between \( y = x^2 \) and \( y = 4 \) from \( x = -2 \) to \( x = 2 \): \[ \text{Area} = \int_{-2}^2 (4 - x^2) \, dx. \] Since \( 4 - x^2 \) is even: \[ \text{Area} = 2 \int_0^2 (4 - x^2) \, dx = 2 \left[ 4x - \frac{x^3}{3} \right]_0^2 = 2 \left( 4 \cdot 2 - \frac{8}{3} \right) = 2 \cdot \frac{24 - 8}{3} = 2 \cdot \frac{16}{3} = \frac{32}{3}. \] Answer: \( \frac{32}{3} \) square units.