Area Between Curves Calculation Of Y=(x-1)^2 And Y=1/4

by ADMIN 55 views
Iklan Headers

Finding the area enclosed by curves is a fundamental concept in calculus, with applications spanning various fields like physics, engineering, and economics. This article delves into the process of calculating the area bounded by the parabola y = (x - 1)² and the straight line y = 1/4. We will explore the necessary steps, from finding the points of intersection to setting up and evaluating the definite integral. This comprehensive guide aims to provide a clear understanding of the underlying principles and techniques involved in solving such problems. By the end of this article, you'll be equipped with the knowledge and skills to tackle similar area-calculation challenges with confidence.

1. Understanding the Problem: Visualizing the Area

Before diving into the mathematical calculations, it's crucial to visualize the problem. The equation y = (x - 1)² represents a parabola with its vertex at (1, 0) opening upwards. The line y = 1/4 is a horizontal line intersecting the parabola at two points. The area we seek is the region enclosed between these two curves. A clear mental image or a quick sketch can significantly aid in understanding the setup and preventing errors. To accurately visualize the area, consider the following:

  • Parabola: The parabola y = (x - 1)² is a standard upward-opening parabola shifted one unit to the right. Its vertex, the lowest point, is at (1, 0).
  • Horizontal Line: The line y = 1/4 is a horizontal line that intersects the y-axis at 1/4. This line will cut across the parabola, creating the enclosed area we want to calculate.
  • Points of Intersection: The points where the parabola and the line intersect are crucial. These points define the limits of integration for our area calculation. We'll determine these points algebraically in the next section. Understanding the visual representation is key to setting up the integral correctly. We need to ensure that we integrate the difference between the upper and lower functions over the correct interval. Misunderstanding the geometry can lead to setting up the integral incorrectly and thus arriving at a wrong answer. So, taking a moment to sketch or mentally picture the curves can save time and prevent frustration later on. Moreover, visualizing the region helps in anticipating the sign of the integral. Since the area is always a positive quantity, we can use our visual understanding to check if the result of the integration makes sense.

2. Finding the Points of Intersection: Where the Curves Meet

The points of intersection are the key to defining the boundaries of our area. To find them, we need to solve the equations y = (x - 1)² and y = 1/4 simultaneously. This means setting the two expressions for y equal to each other: (x - 1)² = 1/4. Now, we solve for x. Taking the square root of both sides, we get x - 1 = ±1/2. This gives us two possible solutions for x: x = 1 + 1/2 = 3/2 and x = 1 - 1/2 = 1/2. These x-values represent the points where the parabola and the line intersect. To get the complete coordinates, we can plug these x-values back into either equation. Using y = 1/4 (since it's simpler), we get the points of intersection as (1/2, 1/4) and (3/2, 1/4). These points are crucial as they define the limits of integration for calculating the area. Let's break down the steps for finding the points of intersection:

  • Set the equations equal: Equate the expressions for y from both equations. This gives us (x - 1)² = 1/4.
  • Solve for x:
    • Take the square root of both sides: √(x - 1)² = ±√(1/4) which simplifies to x - 1 = ±1/2.
    • Isolate x: Add 1 to both sides, giving x = 1 ± 1/2.
    • Find the two solutions: x₁ = 1 + 1/2 = 3/2 and x₂ = 1 - 1/2 = 1/2.
  • Find the y-coordinates: Substitute the x-values back into either of the original equations to find the corresponding y-values. Using y = 1/4 is straightforward.
    • For x = 3/2, y = 1/4.
    • For x = 1/2, y = 1/4.
  • Points of Intersection: The points of intersection are therefore (1/2, 1/4) and (3/2, 1/4).

Identifying the correct points of intersection is paramount. An error here will propagate through the rest of the calculation, leading to an incorrect result. It's always a good practice to double-check these points, perhaps by substituting them back into both original equations to confirm they satisfy both. These points not only define the limits of integration but also help in determining which function is above the other within the interval of integration. This is crucial for setting up the integral correctly, as we need to subtract the lower function from the upper function.

3. Setting Up the Integral: Defining the Area as a Definite Integral

Now that we have the points of intersection, we can set up the definite integral to calculate the area. The area bounded by two curves, f(x) and g(x), between x = a and x = b, where f(x) ≥ g(x) on the interval [a, b], is given by the integral: ∫[a, b] (f(x) - g(x)) dx. In our case, the line y = 1/4 is above the parabola y = (x - 1)² within the interval defined by the points of intersection. Therefore, f(x) = 1/4 and g(x) = (x - 1)². Our limits of integration, a and b, are the x-coordinates of the points of intersection, which we found to be 1/2 and 3/2. Thus, the integral representing the area is: ∫[1/2, 3/2] (1/4 - (x - 1)²) dx. The process of setting up the integral is a critical step. It translates the geometric problem of finding an area into an analytical problem of evaluating a definite integral. Let's break down the key components of setting up the integral:

  • Identify the upper and lower functions: Determine which function has larger y-values over the interval of integration. In this case, the line y = 1/4 is above the parabola y = (x - 1)² within the interval [1/2, 3/2]. So, f(x) = 1/4 and g(x) = (x - 1)².
  • Determine the limits of integration: These are the x-coordinates of the points where the curves intersect. We found these to be 1/2 and 3/2.
  • Write the integral: The area is given by the definite integral of the difference between the upper and lower functions, integrated between the limits of integration.
    • Area = ∫[a, b] (f(x) - g(x)) dx
    • In our case, Area = ∫[1/2, 3/2] (1/4 - (x - 1)²) dx

The integrand, (1/4 - (x - 1)²), represents the height of a thin rectangle at each x-value within the interval. The integral sums up the areas of all such rectangles to give the total area bounded by the curves. Correctly identifying the upper and lower functions is vital. Reversing the order of subtraction will result in a negative area, which is geometrically incorrect. Also, ensure that the limits of integration are consistent with the interval where the identified upper and lower functions hold true. If the curves intersect multiple times, the area may need to be calculated in separate segments, with different integrals for each segment.

4. Evaluating the Integral: Finding the Numerical Value of the Area

Now that we have the integral set up, the next step is to evaluate it. This involves finding the antiderivative of the integrand and then applying the Fundamental Theorem of Calculus. Our integral is: ∫[1/2, 3/2] (1/4 - (x - 1)²) dx. First, let's find the antiderivative of (1/4 - (x - 1)²). Expanding the square, we get 1/4 - (x² - 2x + 1) = 1/4 - x² + 2x - 1 = -x² + 2x - 3/4. Now, we find the antiderivative: ∫(-x² + 2x - 3/4) dx = -x³/3 + x² - (3/4)x + C. Next, we apply the Fundamental Theorem of Calculus, which states that ∫[a, b] f(x) dx = F(b) - F(a), where F(x) is the antiderivative of f(x). In our case, a = 1/2, b = 3/2, and F(x) = -x³/3 + x² - (3/4)x. So, we have: F(3/2) = -(3/2)³/3 + (3/2)² - (3/4)(3/2) = -27/24 + 9/4 - 9/8 = -9/8 + 18/8 - 9/8 = 0. F(1/2) = -(1/2)³/3 + (1/2)² - (3/4)(1/2) = -1/24 + 1/4 - 3/8 = -1/24 + 6/24 - 9/24 = -4/24 = -1/6. Therefore, the area is F(3/2) - F(1/2) = 0 - (-1/6) = 1/6. So, the area bounded by the curves y = (x - 1)² and y = 1/4 is 1/6 square units. Let's break down the steps for evaluating the integral:

  • Expand and simplify the integrand:
    • 1/4 - (x - 1)² = 1/4 - (x² - 2x + 1) = 1/4 - x² + 2x - 1 = -x² + 2x - 3/4
  • Find the antiderivative:
    • ∫(-x² + 2x - 3/4) dx = -x³/3 + x² - (3/4)x + C (We can ignore the constant of integration, C, for definite integrals)
  • Apply the Fundamental Theorem of Calculus:
    • Evaluate the antiderivative at the upper and lower limits of integration and subtract the results.
    • F(x) = -x³/3 + x² - (3/4)x
    • F(3/2) = -(3/2)³/3 + (3/2)² - (3/4)(3/2) = -27/24 + 9/4 - 9/8 = -9/8 + 18/8 - 9/8 = 0
    • F(1/2) = -(1/2)³/3 + (1/2)² - (3/4)(1/2) = -1/24 + 1/4 - 3/8 = -1/24 + 6/24 - 9/24 = -4/24 = -1/6
    • Area = F(3/2) - F(1/2) = 0 - (-1/6) = 1/6

Double-checking the integration and the arithmetic calculations is essential. A small error in any step can lead to an incorrect final answer. Once the final area is obtained, it's good practice to reflect on the result and see if it makes sense in the context of the original problem. The area should be a positive value, and its magnitude should be reasonable given the geometry of the curves.

5. Conclusion: The Area and Its Significance

We have successfully calculated the area bounded by the parabola y = (x - 1)² and the line y = 1/4 to be 1/6 square units. This process involved several key steps: visualizing the problem, finding the points of intersection, setting up the definite integral, and evaluating the integral. Each step is crucial, and a thorough understanding of each is essential for solving similar problems. The area calculated represents the region enclosed between the two curves. It's a positive quantity that measures the two-dimensional space bounded by the parabola and the line. The significance of this calculation extends beyond the realm of pure mathematics. Area calculations are fundamental in various applications:

  • Physics: Calculating areas under velocity-time curves to find displacement, or areas in phase space in statistical mechanics.
  • Engineering: Determining cross-sectional areas for structural analysis, or areas in fluid dynamics calculations.
  • Economics: Finding consumer surplus and producer surplus in market analysis.
  • Probability and Statistics: Calculating probabilities associated with continuous random variables by finding areas under probability density functions.

The techniques and concepts discussed in this article are applicable to a wide range of problems involving area calculations. The key is to understand the underlying principles and apply them systematically. Let's recap the main takeaways from our journey:

  • Visualization: Always start by visualizing the curves and the area you want to calculate. A sketch can be invaluable.
  • Points of Intersection: Finding these points accurately is critical for setting up the integral correctly.
  • Setting up the Integral: Identify the upper and lower functions and the limits of integration carefully.
  • Evaluation: Pay close attention to detail while evaluating the integral to avoid arithmetic errors.
  • Interpretation: Once you have the area, think about what it represents and whether the answer makes sense.

By mastering these techniques, you'll be well-equipped to tackle a wide range of area-calculation problems. Remember that practice is key, so work through various examples to solidify your understanding and build confidence. The journey through calculus is a rewarding one, and the ability to calculate areas is a powerful tool in your mathematical arsenal.

Practice Problems

To further enhance your understanding, here are a few practice problems related to area calculation:

  1. Find the area bounded by the curves y = x² and y = 2x.
  2. Calculate the area enclosed by the curves y = sin(x) and y = cos(x) between x = 0 and x = π/2.
  3. Determine the area between the curve y = x³ - x and the x-axis.

By working through these problems, you can reinforce your understanding of the concepts and techniques discussed in this article. Remember to follow the steps outlined earlier: visualize the problem, find the points of intersection, set up the integral, and evaluate it carefully.

Final Thoughts

Area calculation is a fundamental concept in calculus with far-reaching applications. By understanding the principles and techniques involved, you can unlock a powerful tool for solving problems in various fields. We hope this comprehensive guide has provided you with a clear understanding of how to calculate the area bounded by curves. Keep practicing, keep exploring, and keep expanding your mathematical horizons!