Question

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

Hint:

To find the area between a curve and a line, first determine the intersection points and then integrate the difference between the functions.

Answer 1

Step 1: Set up the equation.
The area of the region is bounded by the curve \( y = x^2 \) and the line \( y = 4 \). To find the limits of integration, we equate the curve and the line: \[ x^2 = 4 \Rightarrow x = \pm 2. \] Thus, the area will be calculated between \( x = -2 \) and \( x = 2 \).

Step 2: Find the area using integration.
The area between the curve and the line is given by the integral of the difference between the line and the curve: \[ \text{Area} = \int_{-2}^{2} (4 - x^2) \, dx \]

Step 3: Solve the integral.
We can now evaluate the integral: \[ \text{Area} = \int_{-2}^{2} 4 \, dx - \int_{-2}^{2} x^2 \, dx \] The first integral is straightforward: \[ \int_{-2}^{2} 4 \, dx = 4x \Big|_{-2}^{2} = 4(2) - 4(-2) = 16. \] The second integral is: \[ \int_{-2}^{2} x^2 \, dx = \frac{x^3}{3} \Big|_{-2}^{2} = \frac{(2)^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \frac{-8}{3} = \frac{16}{3}. \] Thus, the area is: \[ \text{Area} = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3}. \]

Step 4: Conclusion.
The area of the region bounded by the curve \( y = x^2 \) and the line \( y = 4 \) is \( \frac{32}{3} \) square units.