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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.
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00:32
Frank Lin
00:50
Amrita Bhasin
01:40
Linda Hand
Calculus 1 / AB
Chapter 5
Integrals
Section 4
Indefinite Integrals and the Net Change Theorem
Integration
Missouri State University
Campbell University
Harvey Mudd College
University of Michigan - Ann Arbor
Lectures
05:53
In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.
40:35
In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.
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The area of the region tha…
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The area of the regio…
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Represent the area of the …
Find the area of the regio…
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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
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