To find the area bounded by the curves \( y = 4 - x^2 \) and \( y = 2x + 1 \), we will follow a systematic approach. Here’s a detailed step-by-step explanation of how to solve this problem. ### Step 1: Find the Points of Intersection First, we need to determine where the two curves intersect. This is done by setting the equations equal to each other: \[ 4 - x^2 = 2x + 1 \] Rearranging this equation gives: \[ -x^2 - 2x + 4 - 1 = 0 \] \[ -x^2 - 2x + 3 = 0 \] Multiplying through by -1 to make the leading coefficient positive: \[ x^2 + 2x - 3 = 0 \] Next, we can factor this quadratic equation: \[ (x + 3)(x - 1) = 0 \] Setting each factor to zero gives us the solutions: \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] Thus, the points of intersection are \( x = -3 \) and \( x = 1 \). ### Step 2: Set Up the Integral for Area The area \( A \) between the two curves from \( x = -3 \) to \( x = 1 \) can be found using the integral: \[ A = \int_{-3}^{1} \left( (4 - x^2) - (2x + 1) \right) \, dx \] This simplifies to: \[ A = \int_{-3}^{1} \left( 4 - x^2 - 2x - 1 \right) \, dx \] \[ A = \int_{-3}^{1} \left( 3 - x^2 - 2x \right) \, dx \] ### Step 3: Calculate the Integral Now we will compute the integral: \[ A = \int_{-3}^{1} (3 - 2x - x^2) \, dx \] We can integrate term by term: 1. The integral of \( 3 \) is \( 3x \). 2. The integral of \( -2x \) is \( -x^2 \). 3. The integral of \( -x^2 \) is \( -\frac{x^3}{3} \). Thus, we have: \[ A = \left[ 3x - x^2 - \frac{x^3}{3} \right]_{-3}^{1} \] Now we evaluate this expression at the bounds: 1. At \( x = 1 \): \[ 3(1) - (1)^2 - \frac{(1)^3}{3} = 3 - 1 - \frac{1}{3} = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \] 2. At \( x = -3 \): \[ 3(-3) - (-3)^2 - \frac{(-3)^3}{3} = -9 - 9 + 9 = -9 \] Now, substituting these values back into the area formula: \[ A = \left( \frac{5}{3} - (-9) \right) = \frac{5}{3} + 9 = \frac{5}{3} + \frac{27}{3} = \frac{32}{3} \] ### Step 4: Final Area Calculation The area \( A \) is: \[ A = \frac{32}{3} \text{ square units} \] ### Step 5: Review the Options Now, let’s compare this result with the provided options: - A. \( 20(1/3) \) sq. units = \( \frac{20}{3} \) - B. \( 20(2/3) \) sq. units = \( \frac{40}{3} \) - C. \( 10(2/3) \) sq. units = \( \frac{20}{3} \) - D. \( 10(1/3) \) sq. units = \( \frac{10}{3} \) None of the options match \( \frac{32}{3} \). It appears there may have been a misunderstanding in the options provided or the calculations. ### Summary of Key Points - **Intersection Points**: Found by setting the equations equal and solving for \( x \). - **Area Calculation**: Set up the integral of the difference between the two curves. - **Integral Evaluation**: Carefully compute the definite integral to find the area. - **Final Result**: The area bounded by the curves is \( \frac{32}{3} \) square units, which does not match the provided options. ### Revision Summary - Identify points of intersection by solving the equations. - Set up the area integral as the difference of the two functions. - Evaluate the integral carefully, ensuring all calculations are correct. - Double-check the final answer against provided options for accuracy.