Consider a region R=(x,y)∈ R^2:x^2≤ y≤ 2x. If a line y=α divides the area of region R into two equal parts, then which of the following is true ?

$${\int }_{0}^{4}(\sqrt{y}-\frac{y}{2})dy=2{\int }_{0}^{\alpha }(\sqrt{y}-\frac{y}{2})dy$$∫04​(y​−2y​)dy=2∫0α​(y​−2y​)dy

$${(\frac{2{y}^{3/2}}{3}-\frac{{y}^{2}}{2})}_{0}^{4}={2(\frac{2{y}^{3/2}}{3}-\frac{{y}^{2}}{4})}_{0}^{\alpha }$$(32y3/2​−2y2​)04​=2(32y3/2​−4y2​)0α​

$$\Rightarrow \frac{16}{3}-4=2(\frac{2}{3}{\alpha }^{3/2}-\frac{{\alpha }^{2}}{4})$$⇒316​−4=2(32​α3/2−4α2​)

$$\Rightarrow \frac{4}{3}=2(\frac{2}{3}{\alpha }^{3/2}-\frac{{\alpha }^{2}}{4})$$⇒34​=2(32​α3/2−4α2​)

$$\Rightarrow \frac{2}{3}=\frac{8{\alpha }^{3/2}-3{\alpha }^{2}}{12}$$⇒32​=128α3/2−3α2​

$$\Rightarrow 8{\alpha }^{3/2}-3{\alpha }^{2}=8$$⇒8α3/2−3α2=8

$$\Rightarrow 3{\alpha }^{2}-8{\alpha }^{3/2}+8=0$$⇒3α2−8α3/2+8=0