Question

The area of the region \[ \{ (x, y) : x^2 + 4x + 2 \leq y \leq |x| + 2 \} \] is equal to:

• A. 7
• B. \( \frac{24}{5} \)
• C. \( \frac{20}{3} \)
• D. 5

Hint:

To find the area between curves, subtract the lower curve from the upper curve and integrate over the given interval. Ensure that the limits of integration are correctly identified by finding the points of intersection.

Answer 1

The Correct Option is C

Step 1: Rewrite the inequalities. We are given the region defined by \( x^2 + 4x + 2 \le y \le |x+2| \). 
Completing the square in the lower bound gives \( (x+2)^2 - 2 \le y \le |x+2| \).
Step 2: Find the points of intersection of the curves.
We need to find where \( (x+2)^2 - 2 = |x+2| \). Let \( u = x+2 \). Then we have \( u^2 - 2 = |u| \).
Case 1: \( u \ge 0 \). Then \( u^2 - 2 = u \implies u^2 - u - 2 = 0 \implies (u-2)(u+1) = 0 \). Since \( u \ge 0 \), we have \( u = 2 \). Thus \( x + 2 = 2 \implies x = 0 \).
Case 2: \( u<0 \). Then \( u^2 - 2 = -u \implies u^2 + u - 2 = 0 \implies (u+2)(u-1) = 0 \). Since \( u<0 \), we have \( u = -2 \). Thus \( x + 2 = -2 \implies x = -4 \).
So the intersection points are \( x = -4 \) and \( x = 0 \). 
Step 3: Set up the integral for the area. The area of the region is given by the integral: \[ A = \int_{-4}^{0} \left( |x+2| - (x^2 + 4x + 2) \right) dx = \int_{-4}^{0} \left( |x+2| - x^2 - 4x - 2 \right) dx. \] Step 4: Split the integral based on the absolute value. We split the integral into two parts, based on the sign of \( x+2 \). If \( x+2 \ge 0 \), then \( x \ge -2 \), and if \( x+2<0 \), then \( x<-2 \). \[ A = \int_{-4}^{-2} \left( -(x+2) - x^2 - 4x - 2 \right) dx + \int_{-2}^{0} \left( (x+2) - x^2 - 4x - 2 \right) dx. \] \[ A = \int_{-4}^{-2} \left( -x - 2 - x^2 - 4x - 2 \right) dx + \int_{-2}^{0} \left( x + 2 - x^2 - 4x - 2 \right) dx. \] \[ A = \int_{-4}^{-2} \left( -x^2 - 5x - 4 \right) dx + \int_{-2}^{0} \left( -x^2 - 3x \right) dx. \] Step 5: Evaluate the integrals. \[ \int_{-4}^{-2} \left( -x^2 - 5x - 4 \right) dx = \left[ -\frac{x^3}{3} - \frac{5x^2}{2} - 4x \right]_{-4}^{-2} \] \[= \left( \frac{8}{3} - \frac{20}{2} + 8 \right) - \left( \frac{64}{3} - \frac{80}{2} + 16 \right) = \frac{8}{3} - 10 + 8 - \frac{64}{3} + 40 - 16 = - \frac{56}{3} + 22 = \frac{-56+66}{3} = \frac{10}{3}. \] \[ \int_{-2}^{0} \left( -x^2 - 3x \right) dx = \left[ -\frac{x^3}{3} - \frac{3x^2}{2} \right]_{-2}^{0} = 0 - \left( \frac{8}{3} - \frac{12}{2} \right) = - \left( \frac{8}{3} - 6 \right) = - \left( \frac{8-18}{3} \right) = - \left( \frac{-10}{3} \right) = \frac{10}{3}. \] Then \( A = \frac{10}{3} + \frac{10}{3} = \frac{20}{3} \). 
Final Answer: The area of the region is \( \frac{20}{3} \).