Answer is: option3
\( \dfrac{253}{12} \)Solution:
1. Find the Points of Intersection
Set \( f(x) = g(x) \):
\( x + 2 = x^3 - 4x^2 + 6 \)
Rearrange:
\( x^3 - 4x^2 - x + 4 = 0 \)
Factor the equation:
\( (x - 1)(x^2 - 3x - 4) = 0 \Rightarrow (x - 1)(x - 4)(x + 1) = 0 \)
The points of intersection are at \( x = -1, x = 1, \) and \( x = 4 \).
2. Determine Which Function is Above the Other in the Interval
Evaluate \( f(x) \) and \( g(x) \) at a test point within each interval:
- For \( x \in [-1, 1] \), let \( x = 0 \):
\( f(0) = 2, \quad g(0) = 6 \Rightarrow g(x) > f(x) \) - For \( x \in [1, 4] \), let \( x = 2 \):
\( f(2) = 4, \quad g(2) = 2 \Rightarrow f(x) > g(x) \)
3. Set Up the Integrals
The area \( A \) is the sum of the absolute differences between the functions over the intervals where they intersect:
\( A = \int_{-1}^{1} [g(x) - f(x)] \, dx + \int_{1}^{4} [f(x) - g(x)] \, dx \)
Substitute the functions:
\( A = \int_{-1}^{1} (x^3 - 4x^2 - x + 4) \, dx + \int_{1}^{4} (-x^3 + 4x^2 + x - 4) \, dx \)
Add the two results:
\( A = \dfrac{64}{12} + \dfrac{189}{12} = \dfrac{253}{12} \)
Final Answer: \( \boxed{\dfrac{253}{12}} \)
Option C