Calculating the Area Enclosed by a Line, a Parabola, and the X-Axis: A Step-by-Step Guide
Mathematics often becomes a marvel of applications, especially in solving complex problems related to areas and integrations. In this article, we will dive into the detailed calculation of the area enclosed by the line (y 3x), the parabola (y x - 6^2), and the x-axis. This guide will be particularly useful for students, mathematicians, and anyone looking to understand calculus through a practical problem.
Intersection Points
The first step in any calculus problem is identifying the points where the given functions intersect. In this scenario, we are dealing with the line (y 3x) and the parabola (y x - 6^2).
The intersection points are calculated by setting the two equations equal to each other:
[3x x - 6^2]
By rearranging the terms, we have:
[3x - x 6^2]
[2x 36]
[x 18/2 12]
Substituting (x 12) into the line equation, we find (y):
[y 3 cdot 12 36]
Similarly, solving for the other intersection point, we get:
[3x x - 36]
[2x 36]
[x 18/2 3]
Now, substituting (x 3) into the line equation, we find:
[y 3 cdot 3 9]
Thus, our intersection points are (P(12, 36)) and (Q(3, 9)).
Area Between Functions and the X-Axis
To find the area between the line (y 3x) and the x-axis from (x 0) to (x 12), we integrate the function:
[text{Area} int_{0}^{12} 3x , dx]
Evaluating this integral:
[int_{0}^{12} 3x , dx left. frac{3x^2}{2} right|_{0}^{12}]
[ frac{3 cdot 12^2}{2} - frac{3 cdot 0^2}{2}]
[ frac{3 cdot 144}{2} - 0]
[ 216 text{ square units}]
Similarly, for the area between the parabola (y x - 6^2) and the x-axis from (x 0) to (x 6), we integrate:
[text{Area} int_{0}^{6} (x - 6^2) , dx]
Evaluating this integral:
[int_{0}^{6} (x - 72) , dx left. left( frac{x^2}{2} - 72x right) right|_{0}^{6}]
[ left( frac{6^2}{2} - 72 cdot 6 right) - left( frac{0^2}{2} - 72 cdot 0 right)]
[ (18 - 432) - 0]
[ -414]
Since the area cannot be negative, we take the absolute value:
[ 414 text{ square units}]
Combining these results, the total area between the line, the parabola, and the x-axis is:
[216 414 630 text{ square units}]
Integration Between the Functions
To find the enclosed area between the line (y 3x) and the parabola (y x - 6^2), we integrate the difference between these functions:
[text{Area} int_{3}^{12} (3x - (x - 6^2)) , dx]
[ int_{3}^{12} (3x - x 36) , dx]
[ int_{3}^{12} (2x 36) , dx]
Evaluating this integral:
[int_{3}^{12} (2x 36) , dx left. (x^2 36x) right|_{3}^{12}]
[ (12^2 36 cdot 12) - (3^2 36 cdot 3)]
[ (144 432) - (9 108)]
[ 576 - 117]
[ 459 text{ square units}]
Conclusion
Precision in calculating areas within given functions and the x-axis is crucial in calculus, especially in cases where multiple functions overlap. The process involves identifying intersection points, calculating individual areas, and combining differences. This step-by-step guide aims to clarify the process, making it easier to apply similar techniques in future problems.
Keywords
area calculation, parabola, line intersection, calculus application, Google SEO