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

To compute the exact value of the area of the region formed between the graphs of two functions, f(x) = x² + 2x - 6 and g(x) = 2x - 5, we need to follow a systematic approach. This involves finding the points of intersection between the two curves, setting up the integral, and then evaluating it. In this comprehensive guide, we will walk through each step in detail, ensuring a clear understanding of the process. This mathematical problem combines algebraic manipulation with calculus techniques, making it a classic example of how these two branches of mathematics intersect. Understanding how to solve this type of problem is crucial for students studying calculus and for anyone interested in the practical applications of integration.

1. Finding the Points of Intersection

The first step in determining the area between two curves is to find their points of intersection. These points define the limits of integration, indicating where the region begins and ends. To find the points of intersection, we need to set the two functions equal to each other and solve for x. This means we solve the equation f(x) = g(x), which in this case is x² + 2x - 6 = 2x - 5. Simplifying this equation involves rearranging the terms to get a quadratic equation in the standard form ax² + bx + c = 0. By subtracting 2x from both sides and adding 5 to both sides, we get x² - 1 = 0. This quadratic equation is now in a simple form that we can easily solve. We can factor this equation as (x - 1)(x + 1) = 0, which gives us two solutions: x = 1 and x = -1. These are the x-coordinates of the points where the two graphs intersect. To find the corresponding y-coordinates, we can substitute these x-values into either of the original functions. Using g(x) = 2x - 5, when x = 1, g(1) = 2(1) - 5 = -3, and when x = -1, g(-1) = 2(-1) - 5 = -7. Thus, the points of intersection are (1, -3) and (-1, -7). These points are critical because they define the interval over which we will integrate to find the area between the curves.

2. Setting Up the Integral

After identifying the points of intersection, the next step is to set up the definite integral that represents the area between the two curves. The area between two curves, f(x) and g(x), over an interval [a, b] is given by the integral ∫[a, b] |f(x) - g(x)| dx. The absolute value ensures that we are always integrating the positive difference between the two functions, giving us a positive area. In our case, the interval [a, b] is defined by the x-coordinates of the points of intersection, which are -1 and 1. So, we will integrate from -1 to 1. The functions are f(x) = x² + 2x - 6 and g(x) = 2x - 5. We need to determine which function is greater than the other within the interval [-1, 1]. To do this, we can subtract g(x) from f(x): f(x) - g(x) = (x² + 2x - 6) - (2x - 5) = x² - 1. Since x² - 1 is negative within the interval (-1, 1), this means that g(x) is greater than f(x) in this interval. Therefore, the integrand should be |f(x) - g(x)| = |x² - 1| = -(x² - 1) = 1 - x². Now we can set up the definite integral: ∫[-1, 1] (1 - x²) dx. This integral represents the area of the region enclosed between the two curves from x = -1 to x = 1. The setup of the integral is a crucial step, and ensuring the correct order of subtraction and the correct limits of integration is essential for obtaining the correct result. This integral now needs to be evaluated to find the exact value of the area.

3. Evaluating the Integral

With the integral set up, the next step is to evaluate it. We have the definite integral ∫[-1, 1] (1 - x²) dx. To evaluate this integral, we first find the antiderivative of the integrand (1 - x²). The antiderivative of 1 with respect to x is x, and the antiderivative of x² with respect to x is (1/3)x³. Therefore, the antiderivative of (1 - x²) is x - (1/3)x³. Now we need to evaluate this antiderivative at the limits of integration, which are 1 and -1. According to the Fundamental Theorem of Calculus, we subtract the value of the antiderivative at the lower limit from the value at the upper limit. So, we compute [1 - (1/3)(1)³] - [-1 - (1/3)(-1)³]. This simplifies to [1 - 1/3] - [-1 + 1/3], which is 2/3 - (-2/3) = 2/3 + 2/3 = 4/3. Therefore, the exact value of the area of the region formed between the graphs of f(x) = x² + 2x - 6 and g(x) = 2x - 5 is 4/3 square units. This result represents the amount of space enclosed between the two curves in the given interval. The process of evaluating the integral involves applying the basic rules of calculus and algebraic simplification, leading to a precise numerical answer.

4. Visualizing the Region and Verifying the Result

Visualizing the region between the two graphs can help to verify the result and provide a better understanding of the problem. The graphs of f(x) = x² + 2x - 6 and g(x) = 2x - 5 can be sketched on a coordinate plane. The function f(x) is a parabola that opens upwards, and g(x) is a straight line. The region we are interested in is the area enclosed between these two curves between the points of intersection at x = -1 and x = 1. By sketching the graphs, it becomes clear that the region is bounded above by the line g(x) and below by the parabola f(x) in the interval [-1, 1]. The calculated area of 4/3 square units represents the space enclosed within this bounded region. To further verify the result, we can use numerical integration methods or software tools to approximate the area. These methods provide a numerical approximation of the definite integral, which should be close to our calculated value of 4/3. If the numerical approximation is significantly different, it may indicate an error in our calculations, such as an incorrect setup of the integral or a mistake in the integration process. Visualizing the region and using numerical methods to verify the result are valuable steps in ensuring the accuracy of our solution and deepening our understanding of the problem. These steps connect the analytical solution with a geometric interpretation, making the result more intuitive.

5. Alternative Methods and Extensions

While we have calculated the area between the curves using a standard calculus approach, there are alternative methods and extensions to consider. One alternative method involves using symmetry to simplify the calculation. In some cases, the region between the curves may exhibit symmetry about a vertical line. If the functions and the interval of integration are symmetric, we can calculate the area over half the interval and then double the result. However, in this specific case, the region is not symmetric, so this method does not apply. Another extension of this problem involves finding the area between three or more curves. In such cases, we need to identify all the points of intersection and determine the intervals where each function is greater or less than the others. The integral is then set up as a sum of integrals over different intervals, with the appropriate function differences. Additionally, we can explore applications of this concept in various fields. For example, in economics, the area between supply and demand curves represents the consumer surplus or producer surplus. In physics, the area under a velocity-time graph represents the displacement of an object. Understanding how to calculate the area between curves has practical applications in many disciplines. Exploring these alternative methods and extensions can provide a broader perspective on the problem and enhance our problem-solving skills. It also demonstrates the versatility of calculus in addressing real-world problems.

In conclusion, computing the exact value of the area between the graphs of f(x) = x² + 2x - 6 and g(x) = 2x - 5 involves several key steps: finding the points of intersection, setting up the definite integral, evaluating the integral, and verifying the result. This process highlights the interplay between algebra and calculus and provides a solid foundation for solving similar problems. Understanding these concepts is crucial for anyone studying calculus and its applications.