4. Multiple Choice 1 point Find the area of the region bounded by the graphs of the functions \( f(x)=x^{3}-2 x+1 \) and \( g(x)=-2 x \) and \( x=1 \) 2 \( -1 / 4 \) \( 1 / 4 \) 0 Previous

To find the area of the region bounded by the graphs of \( f(x) = x^3 - 2x + 1 \), \( g(x) = -2x \), and the vertical line \( x = 1 \), follow these steps: 1. **Find the Points of Intersection:** - Set \( f(x) = g(x) \): \[ x^3 - 2x + 1 = -2x \implies x^3 + 1 = 0 \implies x = -1 \] - So, the functions intersect at \( x = -1 \). 2. **Set Up the Integral:** - The region of interest is between \( x = -1 \) and \( x = 1 \). - The area \( A \) is given by: \[ A = \int_{-1}^{1} |f(x) - g(x)| \, dx \] - Compute \( f(x) - g(x) \): \[ f(x) - g(x) = (x^3 - 2x + 1) - (-2x) = x^3 + 1 \] - Since \( x^3 + 1 \) is non-negative in the interval \( [-1, 1] \), the absolute value can be removed: \[ A = \int_{-1}^{1} (x^3 + 1) \, dx \] 3. **Evaluate the Integral:** \[ \int_{-1}^{1} x^3 \, dx = \left[\frac{x^4}{4}\right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0 \] \[ \int_{-1}^{1} 1 \, dx = [x]_{-1}^{1} = 1 - (-1) = 2 \] \[ A = 0 + 2 = 2 \] Therefore, the area of the region is **2**. **Answer:** 2