Question

If $f\left(x\right)={\int }_0^1\frac{dx}{\sqrt{4-x^2-x^3}}$ and $a\le f\left(x\right)\le b$

then which of the following is true?

• A. $a=\frac{\pi }{4\sqrt{2}},b=\frac{\pi }{6}$
• B. $a=\frac{\pi }{6},b=\frac{\pi }{4\sqrt{2}}$
• C. $a=\frac{\pi }{4},b=\frac{\pi }{2\sqrt{2}}$
• D. $a=\frac{\pi }{6},b=\frac{\pi }{2\sqrt{2}}$

Answer 1

The correct option is B $a=\frac{\pi }{6},b=\frac{\pi }{4\sqrt{2}}$

As $\sqrt{4-x^2}\ge \sqrt{4-x^2-x^3}\ge \sqrt{4-2x^2}>1\forall xϵ[0,1]$
$\Rightarrow \frac{1}{\sqrt{4-x^2}}\le \frac{1}{\sqrt{4-x^2-x^3}}\le \frac{1}{\sqrt{4-2x^2}}$
$\Rightarrow {\int }_0^1\frac{1}{\sqrt{4-x^2}}dx\le {\int }_0^1\frac{1}{\sqrt{4-x^2-x^3}}dx\le {\int }_0^1\frac{1}{\sqrt{4-2x^2}}dx$
$\Rightarrow {\left[{\sin }^{-1}x\right]}_0^1\le {\int }_0^1\frac{1}{\sqrt{4-x^2-x^3}}dx\le \frac{1}{\sqrt{2}}{\left[{\sin }^{-1}\frac{x}{\sqrt{2}}\right]}_0^1$
$\Rightarrow \frac{\pi }{6}\le {\int }_0^1\frac{1}{\sqrt{4-x^2-x^3}}dx\le \frac{\pi }{4\sqrt{2}}$