Calculate Area Enclosed By Curve Y=x^2-2x-3 And Line Y=x+1
In the realm of calculus, determining the area enclosed between curves is a fundamental concept with wide-ranging applications in various fields such as physics, engineering, and economics. This article delves into the process of calculating the area enclosed between the parabola defined by the equation y = x^2 - 2x - 3 and the straight line given by y = x + 1. We will explore the necessary steps, from finding the points of intersection to setting up and evaluating the definite integral, providing a comprehensive understanding of the underlying principles and techniques involved. Mastering this skill is crucial for anyone seeking to deepen their understanding of calculus and its applications in real-world scenarios.
1. Understanding the Problem
Before embarking on the calculations, it is crucial to visualize the problem and understand the geometric representation of the given equations. The equation y = x^2 - 2x - 3 represents a parabola, a U-shaped curve that opens upwards due to the positive coefficient of the x^2 term. The equation y = x + 1, on the other hand, represents a straight line with a slope of 1 and a y-intercept of 1. To find the area enclosed between these two curves, we need to identify the region where the parabola and the line intersect, forming a closed shape. This region will be bounded by the two curves, and its area can be calculated using integral calculus.
To fully grasp the problem, it's beneficial to sketch the graphs of both the parabola and the line. This visual representation helps in identifying the points of intersection, which are the key to setting up the definite integral for area calculation. By understanding the relative positions of the curves, we can determine which curve lies above the other within the region of interest, a crucial factor in setting up the integral correctly. Moreover, visualizing the problem aids in anticipating the approximate area enclosed, providing a means to verify the final answer obtained through calculations.
2. Finding the Points of Intersection
To determine the area enclosed between the curve and the line, the first critical step involves finding the points of intersection. These points define the boundaries of the region whose area we seek to calculate. Mathematically, the points of intersection are the solutions to the system of equations formed by setting the two equations equal to each other. In this case, we need to solve the equation:
x^2 - 2x - 3 = x + 1
This equation represents the x-coordinates where the y-values of the parabola and the line are equal. To solve it, we rearrange the equation into a standard quadratic form:
x^2 - 3x - 4 = 0
Now, we can solve this quadratic equation by factoring, using the quadratic formula, or completing the square. In this instance, factoring is the most straightforward approach. We look for two numbers that multiply to -4 and add up to -3. These numbers are -4 and 1. Thus, we can factor the quadratic equation as:
(x - 4)(x + 1) = 0
Setting each factor equal to zero gives us the solutions:
x - 4 = 0 => x = 4
x + 1 = 0 => x = -1
These x-values, x = 4 and x = -1, represent the x-coordinates of the points where the parabola and the line intersect. To find the corresponding y-coordinates, we substitute these x-values into either of the original equations. Using the simpler equation, y = x + 1, we get:
For x = 4: y = 4 + 1 = 5
For x = -1: y = -1 + 1 = 0
Therefore, the points of intersection are (-1, 0) and (4, 5). These points define the limits of integration for calculating the area enclosed between the curve and the line.
3. Setting up the Integral
Once we have determined the points of intersection, the next crucial step is to set up the definite integral that represents the area enclosed between the curve and the line. The area between two curves, f(x) and g(x), over an interval [a, b] is given by the integral:
Area = ∫[a, b] |f(x) - g(x)| dx
where f(x) and g(x) are continuous functions, and |f(x) - g(x)| represents the absolute value of the difference between the two functions. The absolute value ensures that we are always integrating a positive difference, as area is a positive quantity. In our case, the functions are f(x) = x + 1 (the line) and g(x) = x^2 - 2x - 3 (the parabola), and the interval is defined by the x-coordinates of the points of intersection, which are a = -1 and b = 4.
Before setting up the integral, it's essential to determine which function lies above the other within the interval of integration. This is because the function on top will have a greater value than the function on the bottom, and we need to subtract the bottom function from the top function to get a positive difference. We can test a point within the interval (-1, 4), such as x = 0, to determine this. Plugging x = 0 into both equations, we get:
f(0) = 0 + 1 = 1
g(0) = 0^2 - 2(0) - 3 = -3
Since f(0) > g(0), the line y = x + 1 lies above the parabola y = x^2 - 2x - 3 in the interval (-1, 4). Therefore, we can set up the integral as:
Area = ∫[-1, 4] [(x + 1) - (x^2 - 2x - 3)] dx
This integral represents the area enclosed between the line and the parabola from x = -1 to x = 4. The next step is to simplify the integrand and evaluate the definite integral.
4. Evaluating the Integral
Having set up the definite integral, the next step is to evaluate it to find the area enclosed between the curve and the line. The integral we need to evaluate is:
Area = ∫[-1, 4] [(x + 1) - (x^2 - 2x - 3)] dx
First, we simplify the integrand by combining like terms:
Area = ∫[-1, 4] (x + 1 - x^2 + 2x + 3) dx
Area = ∫[-1, 4] (-x^2 + 3x + 4) dx
Now, we find the antiderivative of the integrand. Recall that the power rule for integration states that ∫x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. Applying this rule, we find the antiderivative of each term:
∫-x^2 dx = -(x^3)/3
∫3x dx = (3x^2)/2
∫4 dx = 4x
Thus, the antiderivative of -x^2 + 3x + 4 is -(x^3)/3 + (3x^2)/2 + 4x. Now we can evaluate the definite integral using 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, b = 4, and F(x) = -(x^3)/3 + (3x^2)/2 + 4x. Therefore, we need to calculate F(4) - F(-1):
F(4) = -(4^3)/3 + (3(4^2))/2 + 4(4) = -64/3 + 24 + 16 = -64/3 + 40 = (120 - 64)/3 = 56/3
F(-1) = -((-1)^3)/3 + (3(-1)^2)/2 + 4(-1) = 1/3 + 3/2 - 4 = (2 + 9 - 24)/6 = -13/6
Now, we subtract F(-1) from F(4):
Area = F(4) - F(-1) = 56/3 - (-13/6) = 56/3 + 13/6 = (112 + 13)/6 = 125/6
Therefore, the area enclosed between the curve y = x^2 - 2x - 3 and the line y = x + 1 is 125/6 square units. This result represents the magnitude of the region bounded by the parabola and the line, a fundamental concept in calculus with applications in various fields.
5. Conclusion
In conclusion, finding the area enclosed between the curve y = x^2 - 2x - 3 and the line y = x + 1 involves a systematic approach rooted in the principles of integral calculus. The key steps include identifying the points of intersection, setting up the definite integral by determining which function lies above the other, and evaluating the integral using the Fundamental Theorem of Calculus. This process not only provides a numerical value for the area but also reinforces the understanding of how calculus connects algebraic equations with geometric concepts.
The area enclosed, calculated to be 125/6 square units, represents the magnitude of the region bounded by the parabola and the line. This concept has far-reaching applications in various fields, including physics, where it can be used to calculate work done by a force, and economics, where it can represent consumer surplus. By mastering this technique, one gains a valuable tool for solving a wide range of problems that involve areas bounded by curves. Furthermore, the process of setting up and evaluating integrals enhances problem-solving skills, analytical thinking, and the ability to apply mathematical concepts to real-world scenarios.
Therefore, understanding and practicing the methods for finding areas enclosed between curves is essential for students and professionals alike. It forms a cornerstone of calculus knowledge and provides a foundation for more advanced mathematical studies and applications. The ability to visualize the problem, set up the integral correctly, and evaluate it accurately is a testament to one's understanding of calculus and its power in solving geometric and practical problems.
