Here's a definite integral whose value carries memories of grade school. Is there a useful generalization ?
$$ \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \ dx = \frac{22}{7} - \pi$$
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Indeed, there is. – Potato Sep 08 '13 at 21:32
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1Is there an integral that proves $\pi > 333/106$? – Zev Chonoles Sep 08 '13 at 21:33
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See also http://math.stackexchange.com/questions/1079024/expression-for-int-01-xn1-xn-1x2-dx – Simon S Feb 14 '16 at 07:17
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On 28 November 2003 I created this Wikipedia article, to which various others have contributed since then. I think I was the one who added some generalizations, but I'm not sure.
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I'm not sure that would be difficult to do, but if you want it to be something that can be seen to be positive without knowing in advance that $355/113>\pi$, that may be harder. For example, certainly $\displaystyle\int_0^1\left(\frac{355}{113}-\pi\right),dx$ $=\dfrac{355}{113}-\pi$, but probably that's not what you're looking for. – Michael Hardy Sep 08 '13 at 21:45
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You aren't far off,I thought it would be nice to have an arctangent to evaluate. Reference: "Integral proofs that 355/113 > pi" Stephen K. Lucas – Alan Sep 08 '13 at 21:56
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Here are seven series and integrals for the first four convergents to $\pi$ and a question to find the missing one.
Series and integrals for inequalities and approximations to $\pi$
A generalization is given by $$\sum_{k=n}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\int_0^1 \frac{x^{4n}(1-x)^4}{1+x^2}dx$$
which allows to compute sequentially closer rational approximations to $\pi$ by adding the terms of the series.
Jaume Oliver Lafont
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