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

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In calculus, determining the area of a region bounded by curves is a fundamental application of integration. This article delves into the process of computing the exact area enclosed between the graphs of two functions: f(x) = x² + 2x - 6 and g(x) = 2x - 5. We will explore the steps involved, including finding the points of intersection, setting up the definite integral, and evaluating the integral to obtain the precise area. Understanding these techniques is crucial for various applications in mathematics, physics, and engineering, where calculating areas of irregular shapes is often necessary.

1. Identifying Intersection Points: The Foundation of Area Calculation

To accurately compute the area between two curves, the first critical step involves identifying the points of intersection. These intersection points define the limits of integration, which are essential for setting up the definite integral. In simpler terms, these points tell us where the two curves meet, marking the boundaries of the region we're interested in. To find these points, we need to solve the equation f(x) = g(x), where f(x) and g(x) are the functions that define the curves.

In our specific case, we have f(x) = x² + 2x - 6 and g(x) = 2x - 5. Setting these two functions equal to each other allows us to find the x-values where the curves intersect:

x² + 2x - 6 = 2x - 5

Now, we simplify the equation by subtracting 2x from both sides and adding 5 to both sides:

x² - 1 = 0

This is a quadratic equation that can be easily solved. We can factor the equation as follows:

(x - 1)(x + 1) = 0

Setting each factor equal to zero gives us the solutions:

x - 1 = 0 => x = 1 x + 1 = 0 => x = -1

Thus, we've found that the two curves intersect at x = -1 and x = 1. These x-values are the x-coordinates of the intersection points. To find the corresponding y-coordinates, we can substitute these x-values into either f(x) or g(x). Let's use g(x) = 2x - 5 for simplicity:

For x = -1: g(-1) = 2(-1) - 5 = -2 - 5 = -7

For x = 1: g(1) = 2(1) - 5 = 2 - 5 = -3

Therefore, the points of intersection are (-1, -7) and (1, -3). These points are crucial because they define the interval over which we will integrate to find the area between the curves. In our subsequent calculations, we will use these intersection points to set the limits of integration, ensuring we compute the area of the precise region bounded by the two curves.

2. Setting Up the Definite Integral: Translating Geometry to Calculus

With the intersection points identified, we can now set up the definite integral that will calculate the area between the curves. The fundamental principle here is that the area between two curves, f(x) and g(x), over an interval [a, b] is given by the integral of the absolute difference between the functions. Mathematically, this is expressed as:

Area = ∫[a, b] |f(x) - g(x)| dx

Where:

  • a and b are the x-coordinates of the intersection points, which we found to be -1 and 1.
  • f(x) and g(x) are the functions defining the curves.
  • The absolute value ensures that we always integrate the positive difference, giving us a positive area.

In our specific problem, f(x) = x² + 2x - 6 and g(x) = 2x - 5, and our interval is [-1, 1]. So, we need to determine which function is greater over this interval. This will allow us to remove the absolute value sign and set up the integral correctly. To figure out which function is on top, we can choose a test point within the interval, say x = 0, and evaluate both functions at this point:

f(0) = (0)² + 2(0) - 6 = -6 g(0) = 2(0) - 5 = -5

Since g(0) > f(0), we can conclude that g(x) is above f(x) over the interval [-1, 1]. This means that the difference g(x) - f(x) will be positive, and we can rewrite the integral without the absolute value sign:

Area = ∫[-1, 1] (g(x) - f(x)) dx

Now, we substitute the functions into the integral:

Area = ∫[-1, 1] ((2x - 5) - (x² + 2x - 6)) dx

Simplifying the integrand, we get:

Area = ∫[-1, 1] (-x² + 1) dx

This definite integral now represents the exact area between the curves f(x) and g(x) over the interval [-1, 1]. The next step is to evaluate this integral, which we will do in the following section. Setting up the integral correctly is a crucial part of solving this problem, as it translates the geometric concept of area into a calculable mathematical expression. By understanding the relationship between the curves and the limits of integration, we can accurately determine the area enclosed between them.

3. Evaluating the Definite Integral: Unveiling the Exact Area

Having successfully set up the definite integral, the next crucial step is to evaluate it. This process involves finding the antiderivative of the integrand and then applying the Fundamental Theorem of Calculus to compute the definite value. Our integral is:

Area = ∫[-1, 1] (-x² + 1) dx

First, we find the antiderivative of -x² + 1. Recall that the power rule for integration states that ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C, where C is the constant of integration. Applying this rule, we find the antiderivative of -x² to be -(x³)/3 and the antiderivative of 1 to be x. Combining these, we get:

Antiderivative = -(x³)/3 + x

Now, we apply the Fundamental Theorem of Calculus, which states that if F(x) is the antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a). In our case, F(x) = -(x³)/3 + x, a = -1, and b = 1. So, we evaluate F(1) and F(-1):

F(1) = -(1³)/3 + 1 = -1/3 + 1 = 2/3 F(-1) = -((-1)³)/3 + (-1) = -(-1)/3 - 1 = 1/3 - 1 = -2/3

Now, we subtract F(-1) from F(1):

Area = F(1) - F(-1) = (2/3) - (-2/3) = 2/3 + 2/3 = 4/3

Therefore, the exact 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 is a precise value, obtained by applying the principles of integral calculus. The process of evaluating the definite integral involves not only finding the antiderivative but also carefully applying the limits of integration to obtain the numerical result. The 4/3 square units represents the geometric area enclosed between the two curves, a fundamental concept in calculus and its applications.

4. Visualizing the Region and the Result

While the numerical result of 4/3 square units provides the exact area, it's often helpful to visualize the region we've calculated. This visual representation can enhance our understanding of the problem and the solution. The region is bounded by the parabola f(x) = x² + 2x - 6 and the line g(x) = 2x - 5 between the points of intersection (-1, -7) and (1, -3).

Imagine plotting these two functions on a graph. The parabola opens upwards, and the line intersects it at the two points we found earlier. The region whose area we calculated is the area enclosed between the parabola and the line. This region is a curved shape, and the integral allowed us to find its area precisely. The value 4/3 represents the number of square units that would fit within this region.

Visualizing the region helps solidify the concept of area between curves and demonstrates how integral calculus provides a powerful tool for calculating areas of irregular shapes. The numerical answer, 4/3, becomes more meaningful when viewed in the context of the graphical representation. This step reinforces the connection between the algebraic solution and the geometric interpretation, a crucial aspect of understanding calculus.

Conclusion: Mastering Area Calculation with Calculus

In conclusion, we have successfully computed the exact area of the region formed between the graphs of f(x) = x² + 2x - 6 and g(x) = 2x - 5. This process involved several key steps, each building upon the previous one:

  1. Identifying the intersection points: We solved the equation f(x) = g(x) to find the x-coordinates of the points where the curves intersect, which defined the limits of integration.
  2. Setting up the definite integral: We determined which function was greater over the interval and set up the integral of the difference between the functions.
  3. Evaluating the definite integral: We found the antiderivative of the integrand and applied the Fundamental Theorem of Calculus to obtain the exact area.
  4. Visualizing the region: We discussed the importance of visualizing the region to enhance understanding of the problem and the solution.

The final result, 4/3 square units, represents the precise area enclosed between the two curves. This exercise demonstrates the power of integral calculus in solving geometric problems and provides a practical example of how these techniques can be applied. Mastering these concepts is crucial for students and professionals in various fields where calculating areas of irregular shapes is necessary. The process of finding the area between curves highlights the fundamental connection between calculus and geometry, showcasing the elegance and utility of mathematical tools in solving real-world problems.