Math Problem Statement

Find the area of the regions enclosed by the line y = 2 and the curve y = x^2 - 2.

Solution

To find the area of the region enclosed by the line y=2y = 2 and the curve y=x22y = x^2 - 2, let's go through the solution step by step.

Step 1: Set up the equations and find intersection points

  1. We have two functions:

    • y=2y = 2 (a horizontal line)
    • y=x22y = x^2 - 2 (a parabola opening upwards)
  2. To find the points of intersection, set these two equations equal to each other: 2=x222 = x^2 - 2 Solving for xx: x2=4    x=±2x^2 = 4 \implies x = \pm 2 So, the points of intersection are at x=2x = -2 and x=2x = 2.

Step 2: Set up the integral for the area

Since the line y=2y = 2 is above the curve y=x22y = x^2 - 2 within the interval x[2,2]x \in [-2, 2], we can find the area between them by integrating the difference between the two functions from x=2x = -2 to x=2x = 2.

Area=22(2(x22))dx\text{Area} = \int_{-2}^{2} \left(2 - (x^2 - 2)\right) \, dx

Simplifying the integrand: =22(4x2)dx= \int_{-2}^{2} \left(4 - x^2\right) \, dx

Step 3: Integrate

Now, integrate 4x24 - x^2 with respect to xx over the interval [2,2][-2, 2]: 22(4x2)dx=224dx22x2dx\int_{-2}^{2} (4 - x^2) \, dx = \int_{-2}^{2} 4 \, dx - \int_{-2}^{2} x^2 \, dx

  1. Integrate 44: 224dx=4x22=4(2)4(2)=8+8=16\int_{-2}^{2} 4 \, dx = 4x \Big|_{-2}^{2} = 4(2) - 4(-2) = 8 + 8 = 16

  2. Integrate x2x^2: 22x2dx=x3322=(2)33(2)33=83(83)=163\int_{-2}^{2} x^2 \, dx = \left. \frac{x^3}{3} \right|_{-2}^{2} = \frac{(2)^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \left(-\frac{8}{3}\right) = \frac{16}{3}

Step 4: Combine the results

The area is: Area=16163=483163=323\text{Area} = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3}

Final Answer:

The area of the region enclosed by the line y=2y = 2 and the curve y=x22y = x^2 - 2 is: 323\frac{32}{3}

Would you like more details or have any questions?


Here are some additional questions related to this topic:

  1. How would the process change if the curve was y=x2+2y = -x^2 + 2?
  2. What if the line was y=3y = 3 instead of y=2y = 2?
  3. How would you find the enclosed area if the parabola opened downward?
  4. Could you use symmetry to simplify the integration process in similar problems?
  5. How does changing the limits of integration affect the calculated area?

Tip: When finding the area between curves, always check which function is above the other over the given interval to set up the integral correctly.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Area Between Curves

Formulas

Area = ∫(f(x) - g(x)) dx over [a, b]

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12