A (allar Qrea entre Curvas \( x=3-4^{2}, x=9+1 \)

To find the area between the curves defined by the equations \( x = 3 - 4y^2 \) and \( x = 9 + y \), we first need to determine the points of intersection of these two curves. 1. **Set the equations equal to each other**: \[ 3 - 4y^2 = 9 + y \] 2. **Rearrange the equation**: \[ -4y^2 - y + 3 - 9 = 0 \] \[ -4y^2 - y - 6 = 0 \] Multiplying through by -1 gives: \[ 4y^2 + y + 6 = 0 \] 3. **Use the quadratic formula**: The quadratic formula is given by: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 4 \), \( b = 1 \), and \( c = 6 \): \[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 4 \cdot 6}}{2 \cdot 4} \] \[ y = \frac{-1 \pm \sqrt{1 - 96}}{8} \] \[ y = \frac{-1 \pm \sqrt{-95}}{8} \] Since the discriminant is negative, there are no real solutions, which means the curves do not intersect. 4. **Determine the area between the curves**: Since the curves do not intersect, we can analyze the area between them over a specific interval. We can choose a range for \( y \) to calculate the area. Let's find the area between the curves from \( y = -1 \) to \( y = 1 \). - For \( y = -1 \): \[ x = 3 - 4(-1)^2 = 3 - 4 = -1 \] \[ x = 9 + (-1) = 8 \] - For \( y = 1 \): \[ x = 3 - 4(1)^2 = 3 - 4 = -1 \] \[ x = 9 + 1 = 10 \] The area \( A \) between the curves from \( y = -1 \) to \( y = 1 \) can be calculated using the integral: \[ A = \int_{-1}^{1} \left( (9 + y) - (3 - 4y^2) \right) \, dy \] Simplifying the integrand: \[ A = \int_{-1}^{1} \left( 9 + y - 3 + 4y^2 \right) \, dy \] \[ A = \int_{-1}^{1} \left( 6 + y + 4y^2 \right) \, dy \] 5. **Calculate the integral**: \[ A = \left[ 6y + \frac{y^2}{2} + \frac{4y^3}{3} \right]_{-1}^{1} \] Evaluating at the bounds: \[ A = \left( 6(1) + \frac{(1)^2}{2} + \frac{4(1)^3}{3} \right) - \left( 6(-1) + \frac{(-1)^2}{2} + \frac{4(-1)^3}{3} \right) \] \[ = \left( 6 + \frac{1}{2} + \frac{4}{3} \right) - \left( -6 + \frac{1}{2} - \frac{4}{3} \right) \] \[ = \left( 6 + \frac{1}{2} + \frac{4}{3} + 6 - \frac{1}{2} + \frac{4}{3} \right) \] \[ = 12 + \frac{8}{3} \] \[ = 12 + 2.67 = 14.67 \] Thus, the area between the curves from \( y = -1 \) to \( y = 1