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Okay, so I need to figure out how to solve this problem: "Find the area of the region bounded by the curves y = x² and y = 4x - x²." Alright, let's start by visualizing these two curves. The first one is a parabola that opens upwards, right? Because the coefficient of x² is positive. Its vertex is at the origin (0,0). The second equation is y = 4x - x². Hmm, that's also a quadratic, but let me rearrange it to standard form. If I rewrite it as y = -x² + 4x, that's a downward-opening parabola because the coefficient of x² is negative. To find the vertex, I can complete the square or use the formula. The x-coordinate of the vertex is at -b/(2a). Here, a = -1 and b = 4, so x = -4/(2*(-1)) = -4/-2 = 2. Plugging x=2 back into the equation, y = 4*2 - (2)^2 = 8 - 4 = 4. So the vertex is at (2,4).
Now, I need to find the region bounded by these two parabolas. To do that, I should find the points where they intersect. That means setting x² equal to 4x - x² and solving for x. Let's set them equal:
x² = 4x - x²
Bring all terms to one side:
x² + x² - 4x = 0
Combine like terms:
2x² - 4x = 0
Factor out 2x:
2x(x - 2) = 0
So, the solutions are x = 0 and x = 2. Therefore, the two curves intersect at x=0 and x=2. Let me confirm the y-values at these points. For x=0: y = 0² = 0, and y = 40 - 0² = 0. So, point (0,0). For x=2: y = 2² = 4, and y = 42 - 2² = 8 - 4 = 4. So, point (2,4). That checks out.
Now, to find the area between these two curves from x=0 to x=2. Since they are both functions of x, I can set up an integral of the top function minus the bottom function over that interval. I need to determine which function is on top between x=0 and x=2.
Let me pick a test point between 0 and 2, say x=1. Plug into both equations:
For y = x²: y = 1.
For y = 4x - x²: y = 4*1 - 1 = 3.
So, at x=1, the second function (y=4x -x²) is on top. Let me check another point, maybe x=0.5:
y = (0.5)^2 = 0.25
y = 4*(0.5) - (0.5)^2 = 2 - 0.25 = 1.75
Again, the second function is higher. How about at x=1.5:
y = (1.5)^2 = 2.25
y = 4*(1.5) - (1.5)^2 = 6 - 2.25 = 3.75
Still, the second function is on top. So between x=0 and x=2, y=4x -x² is above y=x². Therefore, the area should be the integral from 0 to 2 of (4x - x² - x²) dx, which simplifies to the integral of (4x - 2x²) dx from 0 to 2.
Let me write that out:
Area = ∫₀² (4x - 2x²) dx
Now, compute the antiderivative. The integral of 4x is 2x², and the integral of -2x² is -(2/3)x³. So:
Antiderivative = 2x² - (2/3)x³
Evaluate from 0 to 2.
At x=2:
2*(2)^2 - (2/3)(2)^3 = 24 - (2/3)*8 = 8 - 16/3 = 24/3 - 16/3 = 8/3
At x=0:
2*(0)^2 - (2/3)*(0)^3 = 0 - 0 = 0
Subtracting, the area is 8/3 - 0 = 8/3.
Wait, that seems straightforward, but let me double-check. Maybe I made a mistake in setting up the integral or in the antiderivative.
First, confirming the setup: the upper function is indeed 4x -x², lower is x², so the difference is 4x - 2x². The integral limits from 0 to 2 are correct. The antiderivative of 4x is 2x², correct. The antiderivative of -2x² is -(2/3)x³, that's right. Plugging in x=2: 2*(4) = 8, (2/3)*8 = 16/3. So 8 - 16/3 = 24/3 -16/3 = 8/3. Yes, that's correct. At x=0, it's zero, so the total area is 8/3.
Alternatively, maybe I can compute the integral step by step again:
∫₀² (4x - 2x²) dx
= [2x² - (2/3)x³] from 0 to 2
= [2*(2)^2 - (2/3)*(2)^3] - [0 - 0]
= [8 - (16/3)] = (24/3 - 16/3) = 8/3.
Yep, same result. So the area is 8/3 square units.
Alternatively, just to be thorough, maybe I can visualize the region again. The upward parabola and the downward parabola intersecting at (0,0) and (2,4). The area between them is sort of a lens-shaped region. The integral setup seems correct. Another way to check is to compute the integral using another method or verify with geometry if possible, but since it's between two curves, integration is the way to go.
Alternatively, maybe using symmetry or another approach. Let's see, but I don't think there's a simpler geometric formula here. So I think the answer is 8/3. Let me check online if possible, but since I can't, I have to rely on my calculations. Wait, maybe I can compute the integral another way. Let's split it into two integrals:
∫₀² 4x dx - ∫₀² 2x² dx
First integral: 4x from 0 to 2. Antiderivative is 2x², evaluated from 0 to 2: 2*(4) - 0 = 8.
Second integral: 2x² from 0 to 2. Antiderivative is (2/3)x³, evaluated from 0 to 2: (2/3)*8 - 0 = 16/3.
Subtracting: 8 - 16/3 = 24/3 -16/3 = 8/3. Same result. So that's correct.
Therefore, the area is 8/3. I think that's the answer. Let me just recap the steps to make sure I didn't skip anything:
Find points of intersection by setting x² = 4x -x², solve for x: x=0 and x=2.
Determine which function is on top between these points. Test x=1, found that y=4x -x² is higher.
Set up integral of (top - bottom) from 0 to 2: ∫₀² (4x -x² -x²) dx = ∫₀² (4x -2x²) dx.
Compute the integral, get 8/3.
All steps check out. So the final answer should be 8/3.
Final Answer
The area of the region is \boxed{\dfrac{8}{3}}.
To find the area of the region bounded by the curves ( y = x^2 ) and ( y = 4x - x^2 ):
Find the points of intersection:
Set ( x^2 = 4x - x^2 ):
[
x^2 + x^2 - 4x = 0 \implies 2x^2 - 4x = 0 \implies 2x(x - 2) = 0
]
The solutions are ( x = 0 ) and ( x = 2 ). The points of intersection are ( (0, 0) ) and ( (2, 4) ).
Determine which function is on top:
Test a point between ( x = 0 ) and ( x = 2 ), for example ( x = 1 ):
( y = 1^2 = 1 )
( y = 4(1) - 1^2 = 3 )
The function ( y = 4x - x^2 ) is on top.
Set up the integral:
The area is the integral of the top function minus the bottom function from ( x = 0 ) to ( x = 2 ):
[
\text{Area} = \int_{0}^{2} \left( (4x - x^2) - x^2 \right) , dx = \int_{0}^{2} (4x - 2x^2) , dx
]
Compute the integral:
[
\int_{0}^{2} (4x - 2x^2) , dx = \left[ 2x^2 - \frac{2}{3}x^3 \right]_{0}^{2}
]
Evaluate from 0 to 2:
[
\left( 2(2)^2 - \frac{2}{3}(2)^3 \right) - \left( 2(0)^2 - \frac{2}{3}(0)^3 \right) = \left( 8 - \frac{16}{3} \right) - 0 = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}
]
The area of the region is (\boxed{\dfrac{8}{3}}).
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发表于 2025-11-6 09:57:13