Calculating The Area Between Curves F(x) = X² + 2x - 6 And G(x) = 2x - 5

Jul 9, 2025 by qnaunigon 73 views

Calculating the Exact Area Between Curves f(x) = x² + 2x - 6 and g(x) = 2x - 5

Introduction

In the realm of calculus, determining the area between curves stands as a fundamental concept with far-reaching applications. This article delves into the precise calculation of the area enclosed between two specific functions: f(x) = x² + 2x - 6 and g(x) = 2x - 5. We will embark on a step-by-step journey, meticulously outlining the process of finding intersection points, setting up the definite integral, and ultimately evaluating it to arrive at the exact numerical value of the area. This exploration not only reinforces core calculus principles but also showcases the power of mathematical tools in solving geometric problems. Understanding how to compute the area between curves is crucial for various disciplines, including physics, engineering, and economics, where it is used to model and analyze complex systems. For instance, in physics, it can represent the work done by a force, while in economics, it might depict consumer surplus. This article will equip you with the knowledge and skills necessary to tackle similar problems with confidence and precision. We will break down the problem into manageable steps, ensuring that each concept is clearly explained and illustrated. By the end of this discussion, you will have a solid understanding of the theoretical underpinnings and practical applications of this essential calculus technique. This journey through the calculation of the area between curves will not only enhance your mathematical proficiency but also broaden your appreciation for the elegance and utility of calculus in solving real-world challenges. So, let us begin our exploration and unravel the intricacies of this fascinating problem.

1. Finding the Intersection Points

To accurately compute the area nestled between the curves of f(x) = x² + 2x - 6 and g(x) = 2x - 5, the initial and paramount step lies in pinpointing the intersection points. These points, where the two functions meet, delineate the boundaries of the region whose area we seek to determine. Mathematically, the intersection points are the x-values for which f(x) equals g(x). Therefore, our quest begins with solving the equation x² + 2x - 6 = 2x - 5. This equation sets the stage for our algebraic manipulation, guiding us towards the values of x where the curves intersect. Solving this equation involves bringing all terms to one side, resulting in a quadratic equation. Specifically, subtracting 2x from both sides and adding 5 to both sides, we obtain x² - 1 = 0. This simplified quadratic equation is now in a more manageable form, ready for factorization or the application of the quadratic formula. The elegance of this simplification lies in its ability to transform a seemingly complex problem into a straightforward algebraic task. The next step in our journey involves factoring the quadratic equation. Recognizing the difference of squares, we can factor x² - 1 as (x - 1)(x + 1) = 0. This factorization is a crucial step, allowing us to easily identify the roots of the equation, which correspond to the x-coordinates of the intersection points. Setting each factor to zero, we find that x - 1 = 0 gives us x = 1, and x + 1 = 0 gives us x = -1. These two values, x = -1 and x = 1, are the x-coordinates of the points where the curves intersect. These points define the interval over which we will integrate to find the area between the curves. In summary, finding the intersection points is a critical first step in calculating the area between curves. It involves setting the functions equal to each other, simplifying the resulting equation, and solving for x. In this case, we found the intersection points to be at x = -1 and x = 1, which will serve as the limits of integration in the subsequent steps.

2. Setting Up the Definite Integral

With the intersection points firmly established at x = -1 and x = 1, we now transition to the crucial task of setting up the definite integral. This integral is the mathematical expression that will ultimately yield the precise area nestled between the curves of f(x) = x² + 2x - 6 and g(x) = 2x - 5. The fundamental principle behind this setup lies in recognizing that the area between two curves can be conceptualized as the integral of the difference between the functions over the interval defined by the intersection points. To properly set up the integral, we must first determine which function lies above the other within the interval of integration. This is essential because the order of subtraction in the integral directly affects the sign of the result, and we are interested in the positive value of the area. A simple way to ascertain which function is greater is to choose a test value within the interval (-1, 1) and evaluate both functions at that point. For instance, let's take x = 0. At this point, f(0) = 0² + 2(0) - 6 = -6, and g(0) = 2(0) - 5 = -5. Since -5 is greater than -6, we conclude that g(x) lies above f(x) in the interval [-1, 1]. This determination is pivotal because it dictates the order of subtraction within the integral. Now that we know g(x) is greater than f(x) on the interval [-1, 1], we can set up the definite integral as follows: Area = ∫[-1 to 1] (g(x) - f(x)) dx. This integral represents the accumulation of the infinitesimal areas between the curves over the specified interval. Substituting the expressions for f(x) and g(x), we get: Area = ∫[-1 to 1] ((2x - 5) - (x² + 2x - 6)) dx. This is the core of our calculation, the mathematical representation of the area we seek. Before we proceed to evaluate this integral, it is prudent to simplify the integrand. This simplification not only makes the integration process easier but also reduces the likelihood of errors. Distributing the negative sign and combining like terms, we arrive at: Area = ∫[-1 to 1] (-x² + 1) dx. This simplified integral is now ready for evaluation, the final step in our journey towards determining the exact area between the curves. In summary, setting up the definite integral involves identifying which function is greater within the interval of integration and then expressing the area as the integral of the difference between the functions. The simplification of the integrand is a crucial step that streamlines the subsequent evaluation process.

3. Evaluating the Definite Integral

Having meticulously set up the definite integral, our focus now shifts to the evaluation of this integral, a process that will unveil the precise numerical value of the area enclosed between the curves f(x) = x² + 2x - 6 and g(x) = 2x - 5. The definite integral we've established is: Area = ∫[-1 to 1] (-x² + 1) dx. The evaluation of this integral hinges on the fundamental theorem of calculus, which elegantly connects integration and differentiation. According to this theorem, the definite integral of a function can be found by first determining its antiderivative and then evaluating the antiderivative at the limits of integration. To begin, we seek the antiderivative of -x² + 1. Recall that the power rule for integration states that the antiderivative of x^n is (x^(n+1))/(n+1), provided n ≠ -1. Applying this rule, the antiderivative of -x² is (-x³)/3, and the antiderivative of 1 is x. Therefore, the antiderivative of -x² + 1 is (-x³)/3 + x. This antiderivative is the key to unlocking the value of the definite integral. Now, we evaluate this antiderivative at the limits of integration, x = 1 and x = -1. The fundamental theorem of calculus dictates that we subtract the value of the antiderivative at the lower limit from its value at the upper limit. So, we compute [(-1³)/3 + 1] - [(-(-1)³)/3 + (-1)]. This expression represents the difference in the antiderivative's values at the upper and lower bounds of the integration interval. Simplifying the expression, we have [(-1)/3 + 1] - [(1)/3 - 1]. Further simplification yields (2/3) - (-2/3), which equals 4/3. This numerical result, 4/3, is the exact value of the area between the curves f(x) and g(x) over the interval [-1, 1]. It is a testament to the power of calculus in providing precise solutions to geometric problems. In conclusion, evaluating the definite integral involves finding the antiderivative of the integrand and then applying the fundamental theorem of calculus. This process yields the numerical value of the area, which in this case is 4/3 square units. This result not only answers the specific question at hand but also reinforces the broader applicability of calculus in quantifying geometric concepts.

Conclusion

In this comprehensive exploration, we have successfully navigated the process of calculating the exact area nestled between the curves of f(x) = x² + 2x - 6 and g(x) = 2x - 5. Our journey commenced with the crucial step of pinpointing the intersection points, which we determined to be at x = -1 and x = 1. These points served as the boundaries of the region whose area we sought to quantify. We then transitioned to setting up the definite integral, a mathematical expression that encapsulates the area between the curves. This involved discerning which function lay above the other within the interval of integration and expressing the area as the integral of the difference between the functions. The simplification of the integrand was a key step in streamlining the subsequent evaluation process. Finally, we embarked on the evaluation of the definite integral, a process rooted in the fundamental theorem of calculus. By finding the antiderivative of the integrand and applying the theorem, we arrived at the precise numerical value of the area, which we found to be 4/3 square units. This result not only provides a concrete answer to the problem at hand but also underscores the power and elegance of calculus in solving geometric challenges. The techniques and concepts explored in this article have far-reaching applications in various fields, including physics, engineering, and economics. The ability to calculate the area between curves is essential for modeling and analyzing complex systems, making it a valuable tool in the arsenal of any mathematician or scientist. Furthermore, this exploration serves as a testament to the interconnectedness of mathematical concepts. From algebraic manipulation to the fundamental theorem of calculus, each step in the process builds upon previous knowledge, highlighting the cohesive nature of mathematics. As we conclude this discussion, it is important to recognize that the journey of mathematical exploration is an ongoing one. There are always new problems to solve, new concepts to explore, and new connections to uncover. By embracing this spirit of inquiry, we can continue to deepen our understanding of the mathematical world and its applications.