Question

If the line $x=\alpha$ divides the area of region $R=\left\{(x, y) \in R ^{2}: x^{3} \leq y \leq x, 0 \leq x \leq 1\right\}$ 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: (B), (C)

$\frac{1}{2}\left(\int\limits_{0}^{1}\left( x - x ^{3}\right) dx \right)=\int\limits_{0}^{\alpha}\left( x - x ^{3}\right) dx$
$\frac{1}{8}=\left|\frac{ x ^{2}}{2}-\frac{ x ^{4}}{4}\right|_{0}^{\alpha}$
$\frac{1}{2}=2 \alpha^{2}-\alpha^{4}$
$2 \alpha^{4}-4 \alpha^{2}+1=0$
$f (\alpha)=2 \alpha^{4}-4 \alpha^{2}+1$