Question

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

• A. \( \frac{343}{6} \)
• B. \( \frac{637}{6} \)
• C. \( \frac{437}{6} \)
• D. \( \frac{523}{6} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Question
We need to find the area of a region bounded by two curves: a parabola \( y = x^2 - 8x \) and a V-shaped curve \( y = -|x| \). The inequality \( x^2 - 8x \le y \le -|x| \) means the region is above the parabola and below the V-shaped curve.
Step 2: Key Formula or Approach
1. Identify the bounding curves: upper curve \( y_{upper} \) and lower curve \( y_{lower} \).
2. Find the points of intersection of the curves to determine the limits of integration.
3. The area \(A\) between two curves from \(x=a\) to \(x=b\) is given by the integral: \[ A = \int_{a}^{b} (y_{upper}(x) - y_{lower}(x)) dx \] 4. Handle the absolute value function by splitting the problem into cases: \( -|x| = -x \) for \(x \ge 0\) and \( -|x| = x \) for \(x<0\).
Step 3: Detailed Explanation
The bounding curves are \( y = x^2 - 8x \) (a parabola opening upwards) and \( y = -|x| \).
We need to find the intersection points.
Case 1: \( x \ge 0 \)
The V-shaped curve is \( y = -x \). We find the intersection with the parabola: \[ x^2 - 8x = -x \] \[ x^2 - 7x = 0 \] \[ x(x - 7) = 0 \] The intersection points are at \(x = 0\) and \(x = 7\).
Case 2: \( x<0 \)
The V-shaped curve is \( y = x \). We find the intersection with the parabola: \[ x^2 - 8x = x \] \[ x^2 - 9x = 0 \] \[ x(x - 9) = 0 \] The solutions are \(x=0\) and \(x=9\). Neither of these satisfies the condition \(x<0\), so there are no intersection points in this region.
The entire bounded region lies between \(x=0\) and \(x=7\). In this interval, \(x \ge 0\), so the upper boundary is \( y_{upper} = -x \) and the lower boundary is \( y_{lower} = x^2 - 8x \).
Now we calculate the area using the definite integral: \[ A = \int_{0}^{7} (y_{upper} - y_{lower}) dx \] \[ A = \int_{0}^{7} ((-x) - (x^2 - 8x)) dx \] \[ A = \int_{0}^{7} (-x - x^2 + 8x) dx \] \[ A = \int_{0}^{7} (7x - x^2) dx \] Now, evaluate the integral: \[ A = \left[ \frac{7x^2}{2} - \frac{x^3}{3} \right]_0^7 \] \[ A = \left( \frac{7(7^2)}{2} - \frac{7^3}{3} \right) - (0) \] \[ A = \frac{7^3}{2} - \frac{7^3}{3} = 343 \left( \frac{1}{2} - \frac{1}{3} \right) \] \[ A = 343 \left( \frac{3 - 2}{6} \right) = 343 \left( \frac{1}{6} \right) = \frac{343}{6} \] Step 4: Final Answer
The area of the region is \( \frac{343}{6} \).