Area Between Curves Calculation Y=x² And Y=4x-x²
In the realm of calculus, finding the area of a region bounded by curves is a classic problem. This article will delve into the process of calculating the area of the region in the xy-plane that is enclosed by the curves y₁ = x² and y₂ = 4x - x². To accomplish this, we will embark on a step-by-step journey, first identifying the points of intersection between the two curves and then employing the power of double integrals to determine the area.
1. Unveiling the Intersection Points
To begin our quest, we must first pinpoint the points where the curves y₁ = x² and y₂ = 4x - x² intersect. These intersection points mark the boundaries of the region whose area we seek to calculate. To find these elusive points, we set the equations of the two curves equal to each other:
x² = 4x - x²
This equation represents the condition where the y-values of both curves are identical. By solving this equation, we will uncover the x-coordinates of the intersection points. Let's rearrange the equation to bring all terms to one side:
2x² - 4x = 0
Now, we can factor out a common factor of 2x:
2x(x - 2) = 0
This equation holds true when either of the factors is equal to zero. Therefore, we have two possible solutions:
2x = 0 or x - 2 = 0
Solving for x in each case, we obtain:
x = 0 or x = 2
These x-values represent the x-coordinates of the intersection points. To find the corresponding y-coordinates, we can substitute these x-values into either of the original curve equations. Let's use y₁ = x²:
For x = 0, y₁ = 0² = 0
For x = 2, y₁ = 2² = 4
Thus, the points of intersection are (0, 0) and (2, 4). These points will serve as the anchors for our integration process.
2. Setting Up the Double Integral
Now that we have identified the boundaries of our region, we can set up the double integral to calculate the area. The area of a region bounded by two curves can be found by integrating the difference between the upper curve and the lower curve over the interval defined by the x-coordinates of the intersection points. In this case, the upper curve is y₂ = 4x - x² and the lower curve is y₁ = x². The interval of integration is from x = 0 to x = 2, as determined by the intersection points.
The double integral for the area is given by:
Area = ∬ dA = ∫₀² ∫x²^(4x-x²) dy dx
This integral represents the summation of infinitesimal areas dA over the region. The inner integral integrates with respect to y, from the lower curve y₁ = x² to the upper curve y₂ = 4x - x². The outer integral then integrates with respect to x, from x = 0 to x = 2. This double integration process effectively sweeps across the region, accumulating the infinitesimal areas to yield the total area.
2.1. Evaluating the Inner Integral
Let's first evaluate the inner integral with respect to y:
∫x²^(4x-x²) dy = [y]x²^(4x-x²) = (4x - x²) - x² = 4x - 2x²
This result represents the vertical length of the region at a given x-value. It is the difference between the y-values of the upper and lower curves at that x-value.
2.2. Evaluating the Outer Integral
Now, we substitute the result of the inner integral into the outer integral and integrate with respect to x:
∫₀² (4x - 2x²) dx = [2x² - (2/3)x³]₀²
We now evaluate the antiderivative at the limits of integration:
[2(2)² - (2/3)(2)³] - [2(0)² - (2/3)(0)³] = [8 - (16/3)] - 0 = 8 - (16/3)
To simplify this expression, we find a common denominator:
(24/3) - (16/3) = 8/3
Therefore, the area of the region bounded by the curves y₁ = x² and y₂ = 4x - x² is 8/3 square units.
3. Conclusion
In this exploration, we have successfully navigated the process of calculating the area of a region bounded by two curves in the xy-plane. We first determined the points of intersection between the curves, which defined the boundaries of our region. Then, we set up a double integral, integrating the difference between the upper and lower curves over the interval defined by the x-coordinates of the intersection points. By evaluating this double integral, we arrived at the solution: the area of the region is 8/3 square units.
This exercise showcases the power of calculus in solving geometric problems. Double integrals provide a versatile tool for calculating areas, volumes, and other properties of regions in multi-dimensional space. By mastering the techniques of integration and understanding the underlying concepts, we can unlock a wide range of applications in mathematics, physics, engineering, and beyond.
Key takeaways:
- Finding intersection points is crucial for defining the limits of integration.
- The double integral represents the summation of infinitesimal areas over the region.
- Evaluating the integral step-by-step ensures accuracy.
The answer is B) 8/3.
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