Question

Consider a region $R=\left\{\right(x,y)\in R^2:x^2\le y\le 2x\}$. If a line $y=\alpha$ divides the area of region $R$ into two equal parts, then which of the following is true

• A. ${\alpha }^3-6{\alpha }^2+16=0$
• B. $3{\alpha }^2-8{\alpha }^{\frac{3}{2}}+8=0$
• C. ${\alpha }^3-6{\alpha }^{\frac{3}{2}}-16$
• D. $3{\alpha }^2-8\alpha +8=0$

Answer 1

The correct option is B $3{\alpha }^2-8{\alpha }^{\frac{3}{2}}+8=0$


Area $={\int }_0^2(2x-x^2)dx={x^2-\frac{x^3}{3}|}_0^2=\frac{4}{3}$
$\therefore {\int }_0^a(\sqrt{y}-\frac{y}{2})dy=\frac{2}{3}$
$\Rightarrow {\frac{2}{3}{y}^{\frac{3}{2}}-\frac{y^2}{4}|}_0^a=\frac{2}{3}$
$\Rightarrow 8\cdot {\alpha }^{\frac{3}{2}}-3{\alpha }^2=8$