Finding area with integration and arbitrary line

zeion
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Homework Statement



Sketch the region bounded by y = x^2 and y = 4. This region is divided into two sub regions of equal area by a line y = c. Find c.

Homework Equations





The Attempt at a Solution



I try to integrate from 0 to a point c and make it equate to from integrating c to 4 like this:

<br /> <br /> \int_{0}^{c} [(\sqrt{y})-(-\sqrt{y})]dy = \int_{c}^{4} [(\sqrt{y})-(-\sqrt{y})]dy<br /> <br />

But after I simplify the c disappears.. is this wrong?
 
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Because of the symmetry of this problem, it suffices to look at the region in the first quadrant only.

\int_{0}^{c} \sqrt{y}~dy = \int_{c}^{4} \sqrt{y}~dy
Try working with this equation - c doesn't disappear.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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