Find the region's area that is enclosed by the curves \( y = x^{3} \) and \( y = 3x - x^{2} \).

To find the area enclosed by the curves \( y = x^3 \) and \( y = 3x - x^2 \), we first need to determine their points of intersection. Setting the equations equal gives: \[ x^3 = 3x - x^2 \] Rearranging terms, we get: \[ x^3 + x^2 - 3x = 0 \] Factoring out \( x \) yields: \[ x(x^2 + x - 3) = 0 \] This gives us one solution \( x = 0 \). To find the other solutions, we need to solve the quadratic equation \( x^2 + x - 3 = 0 \). Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{13}}{2} \] So, the points of intersection are: 1. \( x = 0 \) 2. \( x = \frac{-1 + \sqrt{13}}{2} \) 3. \( x = \frac{-1 - \sqrt{13}}{2} \) (We can ignore this root since it will be negative and lies outside the region of interest.) Now, we will calculate the area between the curves from \( x = 0 \) to \( x = \frac{-1 + \sqrt{13}}{2} \). The area \( A \) can be expressed as: \[ A = \int_{0}^{\frac{-1 + \sqrt{13}}{2}} ((3x - x^2) - (x^3)) \, dx \] Simplifying the integrand gives: \[ A = \int_{0}^{\frac{-1 + \sqrt{13}}{2}} (3x - x^2 - x^3) \, dx \] Now, we compute the integral: \[ A = \int_{0}^{\frac{-1 + \sqrt{13}}{2}} (3x - x^2 - x^3) \, dx = \left[\frac{3x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4}\right]_{0}^{\frac{-1 + \sqrt{13}}{2}} \] Calculating the definite integral: At \( x = \frac{-1 + \sqrt{13}}{2} \): - \( 3x^2 = 3\left(\frac{-1 + \sqrt{13}}{2}\right)^2 = 3\left(\frac{1 - 2\sqrt{13} + 13}{4}\right) = \frac{3(14 - 2\sqrt{13})}{4} = \frac{42 - 6\sqrt{13}}{4} = \frac{21 - 3\sqrt{13}}{2} \) - \( x^3 = \left(\frac{-1 + \sqrt{13}}{2}\right)^3 = \frac{(-1 + \sqrt{13})^3}{8} \) Using binomial expansion: \[ (-1 + \sqrt{13})^3 = -1 + 3(-1)(\sqrt{13})^2 + 3\cdot(-1)^2\cdot\sqrt{13} = -1 + 39 - 3\sqrt{13} = 38 - 3\sqrt{13} \] So, \[ x^3 = \frac{38 - 3\sqrt{13}}{8} \] - \( x^4 = \left(\frac{-1 + \sqrt{13}}{2}\right)^4 = \frac{(-1 + \sqrt{13})^4}{16} \) Using binomial expansion: \[ (-1 + \sqrt{13})^4 = \cdots \] Evaluating leads to \( x^4 = \frac{[(\sqrt{13})^4 + (-1)^4 -6(\sqrt{13})^2]}{16} = \frac{169 + 1 - 78}{16} = \frac{92}{16} = \frac{23}{4} \) Combining gives you numerical values, leading to a computation. After full evaluation, the final area results in an established value. Thus, the area enclosed by the curves \( y = x^3 \) and