To find the area bounded by the curve \( y = x(2-x) \), the x-axis, and the vertical lines \( x = 0 \) and \( x = 2 \), we will follow these steps: ### Step 1: Understand the Curve The equation \( y = x(2-x) \) is a quadratic function. We can rewrite it in standard form: \[ y = -x^2 + 2x \] This is a downward-opening parabola. The roots of the equation (where the curve intersects the x-axis) can be found by setting \( y = 0 \): \[ x(2-x) = 0 \] This gives us \( x = 0 \) and \( x = 2 \). Thus, the curve intersects the x-axis at these points. ### Step 2: Determine the Area The area under the curve from \( x = 0 \) to \( x = 2 \) can be found using definite integration. The area \( A \) is given by: \[ A = \int_{0}^{2} y \, dx = \int_{0}^{2} (x(2-x)) \, dx \] ### Step 3: Set Up the Integral Now we set up the integral: \[ A = \int_{0}^{2} (2x - x^2) \, dx \] ### Step 4: Calculate the Integral We can calculate the integral step by step: 1. **Integrate the function**: \[ \int (2x - x^2) \, dx = x^2 - \frac{x^3}{3} + C \] 2. **Evaluate the definite integral from 0 to 2**: \[ A = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} \] Now, we calculate the upper limit: \[ A = \left(2^2 - \frac{2^3}{3}\right) - \left(0^2 - \frac{0^3}{3}\right) \] \[ = \left(4 - \frac{8}{3}\right) - 0 \] \[ = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \] ### Step 5: Final Answer Thus, the area bounded by the curve, the x-axis, and the lines \( x = 0 \) and \( x = 2 \) is: \[ \boxed{\frac{4}{3}} \text{ square units} \] ### Explanation of Other Options - **Option A (4 sq units)**: This option is incorrect because it does not account for the shape of the parabola and the area under it. The area is not a rectangle or a simple shape that would yield 4 square units. - **Option B (2 sq units)**: This option is also incorrect. While it might seem plausible, the area calculated through integration shows that the area is less than 2 square units. - **Option D (1/3 sq units)**: This option is incorrect as well. The area calculated is significantly larger than this value. ### Common Pitfalls - **Misunderstanding the shape**: Students may visualize the area incorrectly, thinking it is larger than it is due to the curve's height. - **Incorrect integration**: Failing to properly integrate the function or miscalculating the definite integral can lead to wrong answers. - **Forgetting to evaluate at both limits**: Always remember to substitute both limits when calculating definite integrals. ### Revision Summary - The area under the curve \( y = x(2-x) \) from \( x = 0 \) to \( x = 2 \) is found using definite integration. - The correct area is \( \frac{4}{3} \) square units. - Always evaluate the integral at both limits to find the area accurately. - Be cautious of the shape of the curve and ensure proper integration techniques are applied.