Question

If $I={\int }_0^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^{2n}}}$ for $n\ge 1$, the value of $I$ is

• A. less than $1$
• B. more than $\frac{1}{2}$
• C. more than $1$
• D. less than $\frac{1}{2}$

Answer 1

Answer: (A), (B)

The correct options are
A less than $1$
B more than $\frac{1}{2}$

Given : $I={\int }_0^{\frac{1}{2}}\frac{dx}{\sqrt{1+x^{2n}}}forn\ge 1$

For any value of $x\in (1,\frac{1}{2})1<\frac{1}{\sqrt{1-x^{2n}}}<2$

${\int }_0^{\frac{1}{2}}1.dx<{\int }_0^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^{2n}}}<2{\int }_0^{\frac{1}{2}}dx\frac{1}{2}<{\int }_0^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^{2n}}}<1$

Hence the correct answer is $\frac{1}{2}<I<1$