Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 208238 by alcohol last updated on 08/Jun/24

$$\mathrm{S}howthat$$$$\frac{\pi }{4}<\underset{0}{\stackrel{1}{\int }}\sqrt{1-x^4}dxusingx=sint$$$$showthat{\int }_0^1\sqrt{1-x^4}dx<\frac{2\sqrt{2}}{3}$$$$using{\left({\int }_0^{1}f\left(x\right)g\left(x\right)dx\right)}^2<{\int }_0^1{\left(f\left(x\right)\right)}^{2}dx{\int }_0^1{\left(g\left(x\right)\right)}^{2}dx$$

Answered by MM42 last updated on 08/Jun/24

$$case1)$$$$I={\int }_0^1\sqrt{1-x^4}dx={\int }_0^1\sqrt{1-x^2}\sqrt{1+x^2}dx$$$$>{\int }_0^1\sqrt{1-x^2}dx$$$$letx=sint$$$${\Rightarrow I>{\int }_0^{\frac{\pi }{2}}{cos}^{2}tdt=(\frac{1}{2}t+\frac{1}{4}sin2t)]}_0^{\frac{\pi }{2}}$$$$\Rightarrow I>\frac{\pi }{4}✓$$$$case2)$$$$I^2={\left({\int }_0^1\sqrt{1-x^4}dx\right)}^2={\left({\int }_0^1\sqrt{1-x^2}\sqrt{1+x^2}dx\right)}^2$$$$<{\int }_0^1{\left(\sqrt{1-x^2}\right)}^{2}dx\times {\int }_0^1{\left(\sqrt{1+x^2}\right)}^{2}dx$$$$={\int }_0^1(1-x^2)dx\times {\int }_0^1(1+x^2)dx=\frac{8}{9}$$$$\Rightarrow I<\frac{2\sqrt{2}}{3}✓$$$$$$

Commented by alcohol last updated on 08/Jun/24

$$thankyou$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com