Question

The area of the region bounded by the curve \(y=x^2\) and the line \(y=4\) is:

• A. \( \frac{32}{3} \) sq. units
• B. \( \frac{64}{3} \) sq. units
• C. \( \frac{16}{3} \) sq. units
• D. \( 64 \) sq. units

Hint:

When finding the area between curves, always sketch the graphs to identify the upper and lower functions and the region of integration. Utilizing symmetry for even or odd functions can simplify the calculation.

Answer 1

The Correct Option is A

Step 1: Find the points of intersection by setting the equations equal: \( x^2 = 4 \implies x = \pm 2 \). These are the limits of integration.
Step 2: Set up the definite integral for the area. The area is the integral of the upper curve minus the lower curve. Here, \(y=4\) is the upper curve and \(y=x^2\) is the lower curve. \[ A = \int_{-2}^{2} (4 - x^2) dx \] Step 3: Use the property of even functions to simplify the integral. Since \(4-x^2\) is an even function, \( \int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx \). \[ A = 2 \int_{0}^{2} (4 - x^2) dx \] Step 4: Evaluate the integral. \[ A = 2 \left[4x - \frac{x^3}{3}\right]_0^2 = 2 \left( (8 - \frac{8}{3}) - 0 \right) = 2 \left( \frac{16}{3} \right) = \frac{32}{3} \]