Question

If the line $x=\alpha$ divides the area of the region $R=\left\{\right(x,y)\in R^2:x^3\le y\le x,0\le x\le 1\}$ into two equal parts, then

• A. $0<\alpha \le \frac{1}{2}$
• B. $\frac{1}{2}<\alpha <1$
• C. $2{\alpha }^4-4{\alpha }^2+1=0$
• D. ${\alpha }^4+4{\alpha }^2-1=0$

Answer 1

Answer: (C)

$2{\alpha }^4-4{\alpha }^2+1=0$

So, ${\int }_0^{\alpha }(x-x^3)dx={\int }_{\alpha }^1(x-x^3)dx{[\frac{x^2}{2}-\frac{x^4}{4}]}_0^{\alpha }={[\frac{x^2}{2}-\frac{x^4}{4}]}_{\alpha }^1\frac{{\alpha }^2}{2}-\frac{{\alpha }^4}{4}=\frac{1^2}{2}-\frac{1^4}{4}-\frac{{\alpha }^2}{2}+\frac{{\alpha }^4}{4}\frac{{2\alpha }^2}{2}-\frac{2{\alpha }^4}{4}=\frac{1}{4}={\alpha }^2-\frac{{\alpha }^4}{2}\frac{1}{2}={2\alpha }^2-{\alpha }^4{2\alpha }^4-4{\alpha }^2+1=0$

Thus, ${2\alpha }^4-4{\alpha }^2+1=0$