Question

If $I={\int }_0^{1/2}\frac{1}{\sqrt{1-x^{2}n}}dx$, then which one of the following is true?

• A. $I\le \frac{\pi }{6}$
• B. $I\ge \frac{1}{2}$
• C. $I\ge 0$
• D. all of these

Answer 1

The correct option is D all of these

for $n\ge 1$ $-1\le x\le 1$
$1\ge \sqrt{1-x^{2n}}\ge \sqrt{1-x^2}$
$\Rightarrow \frac{1}{\sqrt{1-x^2}}\ge \frac{1}{\sqrt{1-x^{2n}}}\ge 1$
$\Rightarrow {\int }_0^{1/2}\frac{1}{\sqrt{1-x^2}}\ge {\int }_0^{1/2}\frac{1}{\sqrt{1-x^{2n}}}\ge {\int }_0^{1/2}dx$
$\Rightarrow \frac{\pi }{6}\ge {\int }_0^{1/2}\frac{1}{\sqrt{1-x^{2n}}}\ge \frac{1}{2}$