Question

The area (in square units) of the region \(\{(x,y): x^2 - 8x \leq y \leq -x\}\) is:

• A. \( \dfrac{343}{6} \)
• B. \( \dfrac{241}{6} \)
• C. \( \dfrac{221}{6} \)
• D. \( \dfrac{323}{6} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The region is bounded above by the line \( y = -x \) and below by the upward-opening parabola \( y = x^2 - 8x \). To find the area, we first determine the points of intersection of these two curves and then integrate the difference of the functions over that interval.
Step 2: Key Formula or Approach:
1. Find intersection: \( x^2 - 8x = -x \). 2. Area \( A = \int_{a}^{b} [f_{upper}(x) - f_{lower}(x)] \, dx \), where \( a \) and \( b \) are the x-coordinates of the intersection points.
Step 3: Detailed Explanation:
1. Solving for intersection: \[ x^2 - 8x + x = 0 \implies x^2 - 7x = 0 \implies x(x - 7) = 0 \] The points of intersection are \( x = 0 \) and \( x = 7 \). 2. Set up the integral: \[ A = \int_{0}^{7} [(-x) - (x^2 - 8x)] \, dx \] \[ A = \int_{0}^{7} (7x - x^2) \, dx \] 3. Evaluate the integral: \[ A = \left[ \frac{7x^2}{2} - \frac{x^3}{3} \right]_{0}^{7} \] \[ A = \left( \frac{7(49)}{2} - \frac{343}{3} \right) - 0 \] \[ A = \frac{343}{2} - \frac{343}{3} = 343 \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{343}{6} \]
Step 4: Final Answer:
The area of the region is \( \dfrac{343}{6} \) square units.