Question

If $I_1={\int }_0^1\frac{1}{|x|}dx$ and $I_2={\int }_0^1\frac{1}{\sqrt{1+x^2}}dx$, then

• A. $I_1=I_2$
• B. $I_1<I_2$
• C. $I_1>I_2$
• D. $\frac{I_1}{I_2}=2$

Answer 1

The correct option is C $I_1>I_2$

We know that,
$|x|<\sqrt{1+x^2}\Rightarrow \frac{1}{|x|}>\frac{1}{\sqrt{1+x^2}}$
$\therefore {\int }_0^1\frac{1}{|x|}dx>{\int }_0^1\frac{1}{\sqrt{1+x^2}}dx\Rightarrow I_1>I_2$.