prove-0-1-x-4-1-x-4-1-x-2-dx-22-7-pi-

Question Number 104609 by M±th+et+s last updated on 22/Jul/20

p r o v e :

\int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} d x = \frac{22}{7} - π

Answered by Dwaipayan Shikari last updated on 22/Jul/20

\int_{0}^{1} \frac{x^{4} {(1 - 2 x + x^{2})}^{2}}{1 + x^{2}} d x

\int_{0}^{1} \frac{x^{4} ({(1 + x^{2})}^{2} - 4 x (1 + x^{2}) + 4 x^{2})}{1 + x^{2}} d x

\int_{0}^{1} x^{4} (1 + x^{2}) - 4 x^{5} + \frac{4 x^{6}}{1 + x^{2}} d x

\int_{0}^{1} x^{4} + x^{6} - 4 x^{5} + \frac{4 x^{6} + 4}{1 + x^{2}} - \frac{4}{1 + x^{2}} d x

{[\frac{x^{5}}{5} + \frac{x^{7}}{7} - \frac{4 x^{6}}{6}]}_{0}^{1} + \int 4 (x^{4} - x^{2} + 1) - {[4 {t a n}^{- 1} x]}_{0}^{1}

\frac{1}{5} + \frac{1}{7} - \frac{2}{3} + \frac{4}{5} - \frac{4}{3} + 4 - 4 . \frac{π}{4}

= 1 + \frac{1}{7} - 2 + 4 - π

= 3 + \frac{1}{7} - π

= \frac{22}{7} - π

It proves that π is smaller than \frac{22}{7}

Commented by Dwaipayan Shikari last updated on 22/Jul/20

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Commented by Ar Brandon last updated on 22/Jul/20

Great decomposition ��

Commented by M±th+et+s last updated on 22/Jul/20

n i c e w o r k s i r

Commented by Dwaipayan Shikari last updated on 22/Jul/20

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