Question
2. We want to compute the area of the region between the line y = x β 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
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Answer to a math question 2. We want to compute the area of the region between the line y = x β 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
Maude
4.7108 Answers
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 = Β±β(2x + 6).
Step 2: Find the points of intersection by setting the two equations equal to each other:
x - 1 = Β±β(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 Β± β(b^2 - 4ac)) / (2a).
Plugging in the values a = 1, b = -4, and c = 1 into the quadratic formula, we get:
x = (-(-4) Β± β((-4)^2 - 4(1)(1))) / (2(1)).
Simplifying, we have:
x = (4 Β± β(16 - 4)) / 2,
x = (4 Β± β(12)) / 2,
x = (4 Β± 2β(3)) / 2.
Step 4: Simplify the x-values:
x = 2 Β± β(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 - β(3) to x = 2 + β(3), and the second part being from x = 2 + β(3) to x = 2 - β(3).
First part:
β«[(x - 1) - β(2x + 6)] dx from 2 - β(3) to 2 + β(3).
Second part:
β«[β(2x + 6) - (x - 1)] dx from 2 + β(3) to 2 - β(3).
Step 6: Evaluate the integrals using the antiderivative of each function.
First part:
β«[(x - 1) - β(2x + 6)] dx
= β«x - 1 - (2x + 6)^(1/2) dx
= (1/2)x^2 - x - (2/3)(2x + 6)^(3/2)] from 2 - β(3) to 2 + β(3).
Second part:
β«[β(2x + 6) - (x - 1)] dx
= β«(2x + 6)^(1/2) - x + 1 dx
= (2/3)(2x + 6)^(3/2) - (1/2)x^2 + x] from 2 + β(3) to 2 - β(3).
Step 7: Calculate the values of the areas using the evaluated integrals.
First part:
[(1/2)(2 + β(3))^2 - (2 + β(3)) - (2/3)(2(2 + β(3)) + 6)^(3/2)] - [(1/2)(2 - β(3))^2 - (2 - β(3)) - (2/3)(2(2 - β(3)) + 6)^(3/2)].
Second part:
[(2/3)(2(2 - β(3)) + 6)^(3/2) - (1/2)(2 - β(3))^2 + (2 - β(3))] - [(2/3)(2(2 + β(3)) + 6)^(3/2) - (1/2)(2 + β(3))^2 + (2 + β(3))].
After simplifying the expressions, we get the final answer:
Answer: The area of the region between the line y = x β 1 and the parabola y^2 = 2x + 6 is given by the evaluated expressions from the integrals in Step 7.
Step 1: Solve for y in the equation y^2 = 2x + 6.
Taking the square root of both sides, we get:
y = Β±β(2x + 6).
Step 2: Find the points of intersection by setting the two equations equal to each other:
x - 1 = Β±β(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 Β± β(b^2 - 4ac)) / (2a).
Plugging in the values a = 1, b = -4, and c = 1 into the quadratic formula, we get:
x = (-(-4) Β± β((-4)^2 - 4(1)(1))) / (2(1)).
Simplifying, we have:
x = (4 Β± β(16 - 4)) / 2,
x = (4 Β± β(12)) / 2,
x = (4 Β± 2β(3)) / 2.
Step 4: Simplify the x-values:
x = 2 Β± β(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 - β(3) to x = 2 + β(3), and the second part being from x = 2 + β(3) to x = 2 - β(3).
First part:
β«[(x - 1) - β(2x + 6)] dx from 2 - β(3) to 2 + β(3).
Second part:
β«[β(2x + 6) - (x - 1)] dx from 2 + β(3) to 2 - β(3).
Step 6: Evaluate the integrals using the antiderivative of each function.
First part:
β«[(x - 1) - β(2x + 6)] dx
= β«x - 1 - (2x + 6)^(1/2) dx
= (1/2)x^2 - x - (2/3)(2x + 6)^(3/2)] from 2 - β(3) to 2 + β(3).
Second part:
β«[β(2x + 6) - (x - 1)] dx
= β«(2x + 6)^(1/2) - x + 1 dx
= (2/3)(2x + 6)^(3/2) - (1/2)x^2 + x] from 2 + β(3) to 2 - β(3).
Step 7: Calculate the values of the areas using the evaluated integrals.
First part:
[(1/2)(2 + β(3))^2 - (2 + β(3)) - (2/3)(2(2 + β(3)) + 6)^(3/2)] - [(1/2)(2 - β(3))^2 - (2 - β(3)) - (2/3)(2(2 - β(3)) + 6)^(3/2)].
Second part:
[(2/3)(2(2 - β(3)) + 6)^(3/2) - (1/2)(2 - β(3))^2 + (2 - β(3))] - [(2/3)(2(2 + β(3)) + 6)^(3/2) - (1/2)(2 + β(3))^2 + (2 + β(3))].
After simplifying the expressions, we get the final answer:
Answer: The area of the region between the line y = x β 1 and the parabola y^2 = 2x + 6 is given by the evaluated expressions from the integrals in Step 7.
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