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The area of the region that lies to the right of the $ y $-axis and to the left of the parabola $ x = 2y - y^2 $ (the shaded region in the figure) is given by the integral $ \displaystyle \int^2_0 (2y - y^2) \, dy $. (Turn your head clockwise and think of the region as lying below the curve $ x = 2y - y^2 $ from $ y = 0 $ to $ y = 2 $.) Find the area of the region.

$\frac{4}{3}$

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So here we give an example of a certain integral, you have the integral from 0 to 2 of two. Y minus y squared, do we? So we'll just solve this integral to the integral of two. Y is just y squared. And the integral of y squared would be Y to the n plus one over n plus one. So if you're y cube over three and this integral goes from 0 to zero, makes no contribution. Here, Tattoo squared would be four minus eight thirds, so four is 12 3rd, so this becomes equal to 12, 3rd times 8/3 is 4/3. This will be 4/3 units squared. So we also can draw a diagram of this. So when we have a problem of the form of Y times two minus why the X is essentially equivalent to zero when y equals zero And when y equals two. And also since the contribution is minus y squared, the problem is going to be open to the left. In this case. As a result, we have some form of a problem like this. And Manly were interested in finding the area of this region and this gives our final answer