The area enclosed by the curve y^2 = x(2 − x) is given by what definite integral?
Should I begin by square rooting both sides? Not really sure.
y^2=-x^2+2x
y^2 -1 =-(x^2-2x+1)
Y^2+(x-1)^2=1
Isn't that a circle of radius 1, centered at 1,0?
yes, so using symmetry, the area is
a = 4∫[0,1] √(1-(x-1)^2) dx = π
To find the area enclosed by the curve y^2 = x(2 − x), we can make use of definite integrals.
First, let's sketch the curve to get a better idea of its shape. Rewriting the equation in terms of y, we have:
y = ±√(x(2 − x))
By plotting both the positive and negative branches of the square root, we can observe the shape of the curve.
Now, to determine the area enclosed by the curve, we need to find the x-values for which the curve intersects the x-axis. These will be the limits of our definite integral.
Setting y = 0 in the equation y^2 = x(2 − x), we can solve for x:
0 = x(2 − x)
This equation has two solutions: x = 0 and x = 2.
Therefore, the area enclosed by the curve can be found by integrating the curve from x = 0 to x = 2. The definite integral that represents the area is:
∫[0 to 2] √(x(2 − x)) dx
To evaluate this integral, you can use methods such as the power rule or substitution.