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

S h o w t h a t

\frac{π}{4} < \underset{0}{\int^{1}} \sqrt{1 - x^{4}} d x u s i n g x = s i n t

s h o w t h a t \int_{0}^{1} \sqrt{1 - x^{4}} d x < \frac{2 \sqrt{2}}{3}

u s i n g {(\int_{0}^{1} f (x) g (x) d x)}^{2} < \int_{0}^{1} {(f (x))}^{2} d x \int_{0}^{1} {(g (x))}^{2} d x

Answered by MM42 last updated on 08/Jun/24

c a s e 1)

I = \int_{0}^{1} \sqrt{1 - x^{4}} d x = \int_{0}^{1} \sqrt{1 - x^{2}} \sqrt{1 + x^{2}} d x

> \int_{0}^{1} \sqrt{1 - x^{2}} d x

l e t x = s i n t

{\Rightarrow I > \int_{0}^{\frac{π}{2}} {c o s}^{2} t d t = (\frac{1}{2} t + \frac{1}{4} s i n 2 t)]}_{0}^{\frac{π}{2}}

\Rightarrow I > \frac{π}{4} ✓

c a s e 2)

I^{2} = {(\int_{0}^{1} \sqrt{1 - x^{4}} d x)}^{2} = {(\int_{0}^{1} \sqrt{1 - x^{2}} \sqrt{1 + x^{2}} d x)}^{2}

< \int_{0}^{1} {(\sqrt{1 - x^{2}})}^{2} d x \times \int_{0}^{1} {(\sqrt{1 + x^{2}})}^{2} d x

= \int_{0}^{1} (1 - x^{2}) d x \times \int_{0}^{1} (1 + x^{2}) d x = \frac{8}{9}

\Rightarrow I < \frac{2 \sqrt{2}}{3} ✓

Commented by alcohol last updated on 08/Jun/24

t h a n k y o u

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