Question

Let $I={\int }_0^1\frac{dx}{\sqrt{4-x^2-x^3}}$ and $I_1={\int }_0^{1/2}\frac{dx}{\sqrt{1-x^4}}$

Answer 1

As $0\le x\le 1\Rightarrow 4-x^2\ge 4-x^2-x^3\ge 4-2x^2\Rightarrow {\int }_0^1\frac{dx}{\sqrt{4-x^2}}\le I\le {\int }_0^1\frac{dx}{\sqrt{4-2x^2}}$
${\int }_0^1\frac{dx}{\sqrt{4-x^2}}={{\sin }^{-1}\frac{x}{2}|}_0^1=\frac{\pi }{6}{\int }_0^1\frac{dx}{\sqrt{4-2x^2}}=\frac{1}{\sqrt{2}}{\int }_0^1\frac{dx}{\sqrt{2-x^2}}={\frac{1}{\sqrt{2}}{\sin }^{-1}\frac{x}{\sqrt{2}}|}_0^1=\frac{\pi }{4\sqrt{2}}$
And as $0\le x\le 1/2\Rightarrow \sqrt{1-x^2}\le \sqrt{1-x^4}\le 1$
$\Rightarrow \frac{1}{2}={\int }_0^{1/2}1dx\le I_1\le {\int }_0^{1/2}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi }{6}$
A) $I<\frac{\pi }{4\sqrt{2}}<\frac{\pi }{4}<1$
B) $I>\frac{\pi }{6}>\frac{1}{2}>\frac{1}{4}$
C) $I_1<\frac{\pi }{6}<\frac{\pi }{4}<1$
D) $I_1>\frac{1}{2}>\frac{1}{4}$