2. We want to compute the area of the region between the line y = x − 1 and the parabola y2 = 2x + 6 We have two methods to calculate this one. For both methods, start by finding the points of interception and sketching the region. (a) Method 1. Solve for y on the equation y2 = 2x + 6. You will notice that there are two solutions. Look back at the graph. These correspond to the two branches of the parabola when we consider y as a function of x. Try to decompose your area as the sum of two areas that you can write as integrals. Be careful. (b) Method 2. Think of y as the variable and of x as a function of y

To find the area of the region between the line and the parabola, we will use Method 1.

Step 1: Solve for y in the equation y^2 = 2x + 6.
Taking the square root of both sides, we get:
y = ±√(2x + 6).

Step 2: Find the points of intersection by setting the two equations equal to each other:
x - 1 = ±√(2x + 6).

Squaring both sides to remove the square root, we have:
(x - 1)^2 = 2x + 6.

Expanding and rearranging the equation, we get:
x^2 - 4x + 1 = 0.

Step 3: Solve for x by factoring or using the quadratic formula. Since the quadratic equation does not factor easily, we will use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a).

Plugging in the values a = 1, b = -4, and c = 1 into the quadratic formula, we get:
x = (-(-4) ± √((-4)^2 - 4(1)(1))) / (2(1)).
Simplifying, we have:
x = (4 ± √(16 - 4)) / 2,
x = (4 ± √(12)) / 2,
x = (4 ± 2√(3)) / 2.

Step 4: Simplify the x-values:
x = 2 ± √(3).

Step 5: Calculate the areas between the line and the parabola using integrals. We will split the region into two parts, with the first part being from x = 2 - √(3) to x = 2 + √(3), and the second part being from x = 2 + √(3) to x = 2 - √(3).

First part:
∫[(x - 1) - √(2x + 6)] dx from 2 - √(3) to 2 + √(3).

Second part:
∫[√(2x + 6) - (x - 1)] dx from 2 + √(3) to 2 - √(3).

Step 6: Evaluate the integrals using the antiderivative of each function.

First part:
∫[(x - 1) - √(2x + 6)] dx
= ∫x - 1 - (2x + 6)^(1/2) dx
= (1/2)x^2 - x - (2/3)(2x + 6)^(3/2)] from 2 - √(3) to 2 + √(3).

Second part:
∫[√(2x + 6) - (x - 1)] dx
= ∫(2x + 6)^(1/2) - x + 1 dx
= (2/3)(2x + 6)^(3/2) - (1/2)x^2 + x] from 2 + √(3) to 2 - √(3).

Step 7: Calculate the values of the areas using the evaluated integrals.

First part:
[(1/2)(2 + √(3))^2 - (2 + √(3)) - (2/3)(2(2 + √(3)) + 6)^(3/2)] - [(1/2)(2 - √(3))^2 - (2 - √(3)) - (2/3)(2(2 - √(3)) + 6)^(3/2)].

Second part:
[(2/3)(2(2 - √(3)) + 6)^(3/2) - (1/2)(2 - √(3))^2 + (2 - √(3))] - [(2/3)(2(2 + √(3)) + 6)^(3/2) - (1/2)(2 + √(3))^2 + (2 + √(3))].

After simplifying the expressions, we get the final answer:

Answer: The area of the region between the line y = x − 1 and the parabola y^2 = 2x + 6 is given by the evaluated expressions from the integrals in Step 7.