Question:

Find the area of the region enclosed between the curves represented by the equations:

1. Curve 1: y = x^2 - 1
2. Curve 2: y = 2x + 1

Step-by-Step Solution:

To find the area between two curves, we need to determine the points of intersection between the curves, and then integrate the difference of the top curve and bottom curve over that interval.

Step 1: Find the points of intersection:

To find the points of intersection between the curves, we need to set their equations equal to each other:

x^2 - 1 = 2x + 1

Simplifying and re-arranging the equation, we get:

x^2 - 2x - 2 = 0

Using the quadratic formula, we can find the values of x:

x = (-(-2) ± √((-2)^2 - 4(1)(-2))) / (2(1)) x = (2 ± √(4 + 8)) / 2 x = (2 ± √12) / 2 x = 1 ± √3

So, the curves intersect at x = 1 + √3 and x = 1 - √3.

Step 2: Determine the top and bottom curves:

To determine which curve is on top and which one is on the bottom, we can compare their equations at any point within the interval of interest. Let's evaluate y at x = 1:

Curve 1: y = (1)^2 - 1 = 0 Curve 2: y = 2(1) + 1 = 3

From this, we can see that Curve 2 (y = 2x + 1) is on top, and Curve 1 (y = x^2 - 1) is on the bottom.

Step 3: Set up and evaluate the integral for the area:

The area between the curves can be found by integrating the difference between the top curve and bottom curve over the interval [1 - √3, 1 + √3].

The integral representing the area is:

A = ∫[(2x + 1) - (x^2 - 1)] dx

Let's integrate this:

A = ∫(2x + 1 - x^2 + 1) dx A = ∫(-x^2 + 2x + 2) dx

To find the antiderivative of each term, we have:

A = [- (1/3)x^3 + x^2 + 2x] + C

Evaluating the integral from the limits of integration [1 - √3, 1 + √3]:

A = [- (1/3)(1 + √3)^3 + (1 + √3)^2 + 2(1 + √3)] - [- (1/3)(1 - √3)^3 + (1 - √3)^2 + 2(1 - √3)] A = [-(1/3)(1 + 3√3 + 3(√3)^2 + (√3)^3) + (1 + 2√3 + (√3)^2) + 2 + 2√3]

• [-(1/3)(1 - 3√3 + 3(√3)^2 - (√3)^3) + (1 - 2√3 + (√3)^2) + 2 - 2√3]

Simplifying further:

A = [-(1/3)(1 + 3√3 + 9 + 3√3) + (1 + 2√3 + 3) + 2 + 2√3]

• [-(1/3)(1 - 3√3 + 9 - 3√3) + (1 - 2√3 + 3) + 2 - 2√3] A = [-(1/3)(12 + 6√3) + (6 + 2√3) + 2 + 2√3]
• [-(1/3)(10 - 6√3) + (4 - 2√3) + 2 - 2√3] A = [-(4 + 2√3) + (6 + 2√3) + 2 + 2√3]
• [-(10 - 6√3) + (4 - 2√3) + 2 - 2√3] A = -4 - 2√3 + 6 + 2√3 + 2 + 2√3
• (-10 + 6√3 + 4 - 2√3 + 2 - 2√3) A = -4 + 6 + 2 + 10 + 2√3 - 2√3 A = 16

Therefore, the area between the curves is 16 square units.

Answer: The area between the curves y = x^2 - 1 and y = 2x + 1 is 16 square units.