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Question Number 208238 by alcohol last updated on 08/Jun/24
$$\mathrm{S}{how}\:{that} \\ $$$$\frac{\pi}{\mathrm{4}}\:<\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}\:{using}\:{x}\:=\:{sint} \\ $$$${show}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}<\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${using}\:\left(\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){g}\left({x}\right){dx}\right)^{\mathrm{2}} <\int_{\mathrm{0}} ^{\mathrm{1}} \left({f}\left({x}\right)\right)^{\mathrm{2}} {dx}\int_{\mathrm{0}} ^{\mathrm{1}} \left({g}\left({x}\right)\right)^{\mathrm{2}} {dx} \\ $$
Answered by MM42 last updated on 08/Jun/24
![case1) I=∫_0 ^1 (√(1−x^4 ))dx=∫_0 ^1 (√(1−x^2 ))(√(1+x^2 ))dx >∫_0 ^1 (√(1−x^2 ))dx let x=sint ⇒I>∫_0 ^(π/2) cos^2 tdt=((1/2)t+(1/4)sin2t)]_0 ^(π/2) ⇒I>(π/4) ✓ case2) I^2 =(∫_0 ^1 (√(1−x^4 ))dx)^2 =(∫_0 ^1 (√(1−x^2 ))(√(1+x^2 ))dx)^2 <∫_0 ^1 ((√(1−x^2 )))^2 dx×∫_0 ^1 ((√(1+x^2 )))^2 dx =∫_0 ^1 (1−x^2 )dx×∫_0 ^1 (1+x^2 )dx=(8/9) ⇒I<((2(√2))/3) ✓](Q208239.png)
$$\left.{case}\mathrm{1}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$>\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$${let}\:\:{x}={sint} \\ $$$$\left.\Rightarrow{I}>\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}} {tdt}=\left(\frac{\mathrm{1}}{\mathrm{2}}{t}+\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\Rightarrow{I}>\frac{\pi}{\mathrm{4}}\:\:\checkmark \\ $$$$\left.{case}\mathrm{2}\right) \\ $$$${I}^{\mathrm{2}} =\left(\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}\right)^{\mathrm{2}} =\left(\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\right)^{\mathrm{2}} \\ $$$$<\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} {dx}×\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx}×\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx}=\frac{\mathrm{8}}{\mathrm{9}} \\ $$$$\Rightarrow{I}<\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\:\:\:\checkmark \\ $$$$ \\ $$
Commented by alcohol last updated on 08/Jun/24
$${thank}\:{you} \\ $$