Dividing a parabolic region in half.

Let $m$ be the slope of the line. Thus, the line and the parabola intersect where:

$mx = x - x^2$

Solving this for $x$ gives $1 - m$ . (The other solution, $0$ , is extraneous.) The areas of the two regions can thus be expressed by:

$\int_0^{1-m} (x - x^2 - mx)\: dx \quad\quad \text{and} \quad\quad \int_0^{1-m} mx\: dx + \int_{1-m}^1 (x - x^2)\: dx$

Setting these two expressions equal to each other and integrating gives:

$\left.\left(\frac{1-m}{2}x^2-\frac{x^3}{3}\right)\right|_0^{1-m} = \left.\frac{m}{2}x^2\right|_0^{1-m} + \left.\left(\frac{x^2}{2} - \frac{x^3}{3}\right)\right|_{1-m}^1$ $\frac{1-m}{2}(1-m)^2 - \frac{(1-m)^3}{3} = \frac{m}{2}(1-m)^2 + \left(\frac{1}{2} - \frac{1}{3} - \frac{(1-m)^2}{2} + \frac{(1-m)^3}{3}\right)$

Multiplying through by the common denominator $6$ and simplifying gives:

$(1-m)^3 = 3m(1-m)^2 + 3 - 2 - 3(1-m)^2 + 2(1-m)^3$

Further simplification gives the cubic equation $2m^3 - 6m^2 + 6m - 1 = 0$ , whose only real solution is:

$m = 1 - \frac{1}{\sqrt[3]{2}} \approx \boxed{0.206}$