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Prove `int_0^1 1/(1+x^4)dx in [pi /4,1]`

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xalal | Honors

Posted June 27, 2013 at 5:34 PM via web

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Prove `int_0^1 1/(1+x^4)dx in [pi /4,1]`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 27, 2013 at 6:13 PM (Answer #2)

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You need to notice that `x^4 <= x^2` over the interval `[0,1]` such that:

`0 <= x^4 <= x^2 `

Adding 1 in inequality yields:

`1 <= x^4 + 1 <= x^2 + 1 => 1 >= 1/(x^4 + 1) >= 1/(x^2 + 1)`

 

Integrating the inequality yields:

`int_0^1 1 dx >= int_0^1 1/(x^4 + 1) dx >= int_0^1 1/(x^2 + 1) dx`

`x|_0^1 >= int_0^1 1/(x^4 + 1) dx >= tan^(-1) x|_0^1`

Using the fundamental theorem of calculus, yields:

`tan^(-1) 1 - tan^(-1) 0 <= int_0^1 1/(x^4 + 1) dx <= 1 - 0`

`pi/4 - 0 <= int_0^1 1/(x^4 + 1) dx <= 1 => int_0^1 1/(x^4 + 1) dx in [pi/4,1]` .

Hence, testing if the statement ` int_0^1 1/(x^4 + 1) dx` `in [pi/4,1]` is valid, yields that the inequality `pi/4 <= int_0^1 1/(x^4 + 1) dx <= 1.`

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