# Inequality no solve integral Show 0<(equal) integral(0 to 1) fn(x) < (equal)1/2n+1, n natural given fn(x)=x^(2n)/1+x?

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You should notice that if `x in (0,1)`  the following inequality holds such that:

`0 < (x^(2n))/(1 + x) < (x^(2n))/1`

You need to integrate such that:

`int_0^1 0 dx < int_0^1 (x^(2n))/(1 + x) dx < int_0^1 (x^(2n))/1 dx`

`0 < int_0^1 (x^(2n))/(1 + x) dx < (x^(2n+1))/(2n+1)|_0^1`

Using the fundamental theorem of calculus yields:

`0 < int_0^1 (x^(2n))/(1 + x) dx < (1^(2n+1) - 0^(2n+1))/(2n+1)`

`0 < int_0^1 (x^(2n))/(1 + x) dx < (1-0)/(2n+1)`

`0 < int_0^1 (x^(2n))/(1 + x) dx < 1/(2n+1)`

Hence, the inequality may be proved without evaluating the integral, thus `0 < int_0^1f_n(x) dx< 1/(2n+1).`

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