How do you find the area enclosed by the curve #y^2 = x(1 - x)#?
The equation can be reorganized as
(x-1/2)^2+y^2=(1/2)^2 that represents a circle of radius 1/2.
So, the area is #pi(1/2)^2=pi/4 areal units. See the graph.
graph{y^2-x(1-x)=0 [-2.5, 2.5, -1.25, 1.25]}
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To find the area enclosed by the curve ( y^2 = x(1 - x) ), you can integrate ( y ) with respect to ( x ) over the relevant interval of ( x ). First, determine the ( x )-interval by finding the points where the curve intersects the ( x )-axis. Then, integrate the square root of the expression ( x(1 - x) ) from the lowest ( x )-value to the highest. This will give you the area enclosed by the curve.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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