Question

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

• A. \(\frac{343}{6}\)
• B. \(\frac{343}{2}\)
• C. \(\frac{1715}{6}\)
• D. \(\frac{340}{3}\)

Hint:

For the area between \( y = ax^2 + bx + c \) and \( y = mx + k \), if the intersection points are \( \alpha \) and \( \beta \), the area is \( \frac{|a|}{6} (\beta - \alpha)^3 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The region is bounded between the line \( y = -x \) and the parabola \( y = x^2 - 8x \). We find the points of intersection to determine the limits of integration.

Step 2: Key Formula or Approach:
Area = \( \int_{x_1}^{x_2} (y_{upper} - y_{lower}) dx \). Intersection: \( x^2 - 8x = -x \implies x^2 - 7x = 0 \).

Step 3: Detailed Explanation:
1. Limits: \( x = 0 \) and \( x = 7 \). 2. Function difference: On \( [0, 7] \), \( -x \ge x^2 - 8x \). \[ \text{Area} = \int_{0}^{7} (-x - (x^2 - 8x)) dx = \int_{0}^{7} (7x - x^2) dx \] 3. Integration: \[ \text{Area} = \left[ \frac{7x^2}{2} - \frac{x^3}{3} \right]_0^7 = \frac{7(49)}{2} - \frac{343}{3} \] \[ \text{Area} = \frac{343}{2} - \frac{343}{3} = 343 \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{343}{6} \]

Step 4: Final Answer:
The area of the region is \( \frac{343}{6} \) square units.