Question

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

Hint:

To find the area between a curve and a line, first find the points of intersection, then integrate the difference between the curve and the line over the interval.

Answer 1

To find the area between the curve \( y = x^2 \) and the line \( y = 4 \), we first find the points of intersection. Set \( x^2 = 4 \), which gives: \[ x = \pm 2 \] Thus, the points of intersection are \( x = -2 \) and \( x = 2 \). The area is given by the integral: \[ \text{Area} = \int_{-2}^{2} \left( 4 - x^2 \right) \, dx \] Now, compute the integral: \[ \int (4 - x^2) \, dx = 4x - \frac{x^3}{3} \] Substitute the limits: \[ \text{Area} = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} = \left( 4(2) - \frac{(2)^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right) \] Simplifying: \[ \text{Area} = \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right) = 8 - \frac{8}{3} + 8 - \frac{8}{3} = 16 - \frac{16}{3} \] Thus, the area is: \[ \text{Area} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} \] The area of the region is: \[ \boxed{\frac{32}{3}} \]