What is the area between two intersecting parabolas?

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SUMMARY

The area between the intersecting parabolas defined by the equations \(y = x^2\) and \(y = 2x - x^2\) is calculated using definite integrals. The points of intersection are established at \(x = 0\) and \(x = 1\). The correct formula for the area \(A\) is given by \(A = \int_0^1 \left((2x - x^2) - x^2\right) dx\), simplifying to \(A = 2\int_0^1 (x - x^2) dx\). The area must be positive, which is achieved by ensuring the top function is subtracted from the bottom function correctly.

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Find the area of the region enclosed by the parabolas $$y = x^2 $$and $$y = 2x - x^2.$$

So they intersect at 0 and 1.

The derivative of $$y = 2x - x^2.$$ is $$\d{y}{x} = 2 - 2x$$

When I plug in 1 I get 0, and when I plug in 0, I get 2, so I subtract 2 from 0 and the area is -2. But the area should be positive, what am I doing wrong?
 
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The width of each area element is $dx$ and the height is the top function minus the bottom function, thus:

$$A=\int_0^1 \left(2x-x^2\right)-\left(x^2\right)\,dx=2\int_0^1 x-x^2\,dx$$

What do you get?
 

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