Two other approaches (both already exploiting the vertical symmetry so the whole area is just the right half of 2/3): First: compute the bottom half of the area by decomposing into a rectangle of sides c and sqrt(1-c) added to the area under the curve (to the x-axis) between x = c and x = 1. So that's c*sqrt(1-c) + ∫(1-x^2)dx between c and 1. Equate that to 1/3, solve for c. Second: work in terms of strips from the curve to y-axis. So ∫xdy between y = c to y = 1, change variable, ∫xd(1-x^2) = ∫x(-2x)dx = ∫-2x^2dx between x =sqrt(1-c) to x = 0. Again, equate to 1/3, solve for c.
Two other approaches (both already exploiting the vertical symmetry so the whole area is just the right half of 2/3):
First: compute the bottom half of the area by decomposing into a rectangle of sides c and sqrt(1-c) added to the area under the curve (to the x-axis) between x = c and x = 1. So that's c*sqrt(1-c) + ∫(1-x^2)dx between c and 1. Equate that to 1/3, solve for c.
Second: work in terms of strips from the curve to y-axis. So ∫xdy between y = c to y = 1, change variable, ∫xd(1-x^2) = ∫x(-2x)dx = ∫-2x^2dx between x =sqrt(1-c) to x = 0. Again, equate to 1/3, solve for c.
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