Question

The area bounded between two parabolas \( y = x^2 + 4 \) and \( y = -x^2 + 6 \) is

Hint:

When finding the area between two curves, first solve for the points of intersection, then set up and evaluate the integral of the difference of the functions between the intersection points.

Answer 1

Correct Answer: 2.6 - 2.7

Step 1: Find the points of intersection.
To find the points of intersection of the two parabolas, set the equations equal to each other: \[ x^2 + 4 = -x^2 + 6. \] Solving for \( x \), we get: \[ x^2 + x^2 = 6 - 4 \quad \Rightarrow \quad 2x^2 = 2 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1. \] Thus, the points of intersection are at \( x = -1 \) and \( x = 1 \).

Step 2: Set up the integral.
The area between the curves is the integral of the difference between the two functions from \( x = -1 \) to \( x = 1 \): \[ A = \int_{-1}^{1} \left[ (-x^2 + 6) - (x^2 + 4) \right] dx. \] Simplifying the integrand: \[ A = \int_{-1}^{1} \left[ -2x^2 + 2 \right] dx. \]
Step 3: Compute the integral.
Now, evaluate the integral: \[ A = \int_{-1}^{1} (-2x^2 + 2) \, dx. \] The integral of \( -2x^2 \) is \( -\frac{2}{3}x^3 \), and the integral of 2 is \( 2x \). Thus: \[ A = \left[ -\frac{2}{3}x^3 + 2x \right]_{-1}^{1} = \left( -\frac{2}{3}(1)^3 + 2(1) \right) - \left( -\frac{2}{3}(-1)^3 + 2(-1) \right). \] Simplifying: \[ A = \left( -\frac{2}{3} + 2 \right) - \left( \frac{2}{3} - 2 \right) = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}. \] Thus, the area between the curves is \( \frac{8}{3} \approx 2.67 \), which is approximately between 2.6 and 2.7.

Step 4: Conclusion.
Thus, the area between the two parabolas is approximately \( 2.6 \) to \( 2.7 \).