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| Upper bond of Integral http://mathhelpplanet.com/viewtopic.php?f=19&t=22322 |
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| Автор: | jagdish [ 27 фев 2013, 18:21 ] |
| Заголовок сообщения: | Upper bond of Integral |
Prove that [math]\displaystyle 0.78<\int_{0}^{1}\frac{1}{\sqrt{1+x^4}}dx<0.93[/math] My Solution: [math]1+x^4 <(1+x^2)^2\; \forall x \in \left(0,1\right)[/math] So [math]\displaystyle \frac{1}{\sqrt{1+x^4}}> \frac{1}{1+x^2}[/math] [math]\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+x^4}}dx > \int_{0}^{1}\frac{1}{1+x^2}dx = \frac{\pi}{4}[/math] So [math]\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+x^4}}dx>0.78[/math] My Question is How Can I calculate for Upper Bond. Can anyone Like to explain me. Thanks |
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| Автор: | andrei [ 27 фев 2013, 22:32 ] |
| Заголовок сообщения: | Re: Upper bond of Integral |
[math](1+x^{4})^{-0,5}<1- \frac{ 1 }{ 2 }x^{4}+ \frac{ 3 }{ 8 }x^{8} \quad \Rightarrow \quad J<1- \frac{ 1 }{ 10}+ \frac{ 3 }{ 72 }=0,9416...[/math] |
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