If the line x = a divides the area of the region $R=\{(x, y) ∈R^2:x^3 ≤y≤x, 0≤x≤1\}$ into two equal parts, then

$\frac{1}{2} < α <1$

It is given that $x = α$ divides the shaded region into two equal parts.

$∴2\int\limits_0^α(x-x^3)dx=\int\limits_0^1(x-x^3)dx$

$⇒2\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^α=\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1$

$⇒α^2-\frac{α^4}{2}=\frac{1}{4}⇒2α^4-4α^2+1=0$

Let $f(α)=2α^4-4α^2+1$. Then,

$f(1)=-1<0$ and $f(\frac{1}{2})=\frac{1}{8}>0⇒\frac{1}{2}<α<1$