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

The area of the region defined by the inequalities \(x^2 + 4x + 2 \leq y \leq |x| + 2\) is determined through the following steps: 

1. Curve Identification:
   • The function \(y = x^2 + 4x + 2\) represents an upward-opening parabola. Rewriting it by completing the square yields \(y = (x+2)^2 - 2\).
   • The function \(y = |x| + 2\) forms a V-shape, with two linear segments:  \(y = x + 2\) for \(x \geq 0\) and \(y = -x + 2\) for \(x < 0\).
2. Intersection Point Calculation:
   • To find intersections with the right linear segment \(y = x + 2\), we solve: 
   \[x^2 + 4x + 2 = x + 2\]
   • This simplifies to \(x^2 + 3x = 0\), yielding \(x(x+3) = 0\). The intersection points occur at \(x = 0\) and \(x = -3\).
   • To find intersections with the left linear segment \(y = -x + 2\), we solve: 
   \[x^2 + 4x + 2 = -x + 2\]
   • This simplifies to \(x^2 + 5x = 0\), yielding \(x(x+5) = 0\). The intersection points occur at \(x = 0\) and \(x = -5\).
3. Area Computation via Integration:
   • The relevant interval for integration is \(x \in [-5, 0]\). The area is calculated as the sum of two integrals:

\[\text{Area} = \int_{-5}^{-3} ((-x+2) - (x^2 + 4x + 2)) \, dx + \int_{-3}^{0} ((x+2) - (x^2 + 4x + 2)) \, dx\]

• .
• Evaluating the first integral:

\[\begin{align*} \int_{-5}^{-3} (-x^2 - 5x) \, dx &= \left[-\frac{x^3}{3} - \frac{5x^2}{2}\right]_{-5}^{-3} \\ &= \left(-\frac{(-3)^3}{3} - \frac{5(-3)^2}{2}\right) - \left(-\frac{(-5)^3}{3} - \frac{5(-5)^2}{2}\right) \\ &= \left(9 - \frac{45}{2}\right) - \left(\frac{125}{3} - \frac{125}{2}\right) \end{align*}\]

• .
• Evaluating the second integral similarly and summing the results:

\[\begin{align*} \int_{-3}^{0} (-x^2-3x) \, dx &= \left[-\frac{x^3}{3} - \frac{3x^2}{2}\right]_{-3}^{0} \\ &= 0 - \left(-\frac{(-3)^3}{3} - \frac{3 \times (-3)^2}{2}\right) \\ &= \frac{27}{3} + \frac{27}{2} = \frac{81}{6} \end{align*}\]

• .

1. The sum of the two integrals yields the total area: \(\frac{20}{3}\).

Therefore, the area of the specified region is \(\frac{20}{3}\).