Question

The area bounded by the curves \( y = -x^2 + 3 \) and \( y = 0 \) is:

• A. \( \sqrt{3}+1 \)
• B. \( \sqrt{3} \)
• C. \( 4\sqrt{3} \)
• D. \( 5\sqrt{3} \)
• E. \( 6\sqrt{3} \)

Hint:

Area under a parabolic arch from one root to another can also be found using Archimedes' formula: Area = \( \frac{2}{3} \times \text{Base} \times \text{Height} \). Base = \( 2\sqrt{3} \), Height = \( 3 \). Area = \( \frac{2}{3} \cdot 2\sqrt{3} \cdot 3 = 4\sqrt{3} \).

Answer 1

The Correct Option is C

Concept: The area bounded by a curve \( y = f(x) \) and the \( x \)-axis (\( y = 0 \)) is calculated using the definite integral \( \int_{a}^{b} f(x) \, dx \). The limits of integration, \( a \) and \( b \), are the \( x \)-intercepts of the curve where it crosses the \( x \)-axis. These are found by solving \( f(x) = 0 \).

Step 1: Find the limits of integration.
To find where the parabola \( y = -x^2 + 3 \) intersects the \( x \)-axis (\( y = 0 \)), we set: \[ -x^2 + 3 = 0 \implies x^2 = 3 \implies x = \pm\sqrt{3} \] So, the limits are \( a = -\sqrt{3} \) and \( b = \sqrt{3} \).

Step 2: Set up and evaluate the definite integral.
The area is given by: \[ \text{Area} = \int_{-\sqrt{3}}^{\sqrt{3}} (-x^2 + 3) \, dx \] Using the property of even functions (\( \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \)): \[ \text{Area} = 2 \int_{0}^{\sqrt{3}} (-x^2 + 3) \, dx \] \[ \text{Area} = 2 \left[ -\frac{x^3}{3} + 3x \right]_{0}^{\sqrt{3}} = 2 \left[ (-\frac{(\sqrt{3})^3}{3} + 3\sqrt{3}) - 0 \right] \] \[ \text{Area} = 2 \left[ -\frac{3\sqrt{3}}{3} + 3\sqrt{3} \right] = 2 [-\sqrt{3} + 3\sqrt{3}] = 2 [2\sqrt{3}] = 4\sqrt{3} \]