# What is definite integral between -1, 1 in y=1 over (e to (x)+1)(x to (2)+1)?explain how is function

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find if the function is odd or even such that:  if the function is odd then f(-x)=-f(x) and if the function is even then f(-x)=f(x).

Calculating f(-x) yields:

`f(-x) = 1/(e^(-x) + 1)((-x)^2 + 1) =gt f(-x) = 1/(1/e^x + 1)(x^2 + 1)`

Notice that the equation of the function f(-x) does not express either f(x), or -f(x), hence the function is neither even, nor odd.

You need to remember the formula for symmetric definite integral:a

`int_-a^a f(x) dx= int_0^a [f(x) + f(-x)]dx`

Hence `int_-1^1 f(x) dx= int_0^1 [f(x) + f(-x)]dx`

Calculating `[f(x) + f(-x)] ` yields:

`[f(x) + f(-x)] = 1/((e^x + 1)(x^2 + 1)) + 1/(e^(-x) + 1)((-x)^2 + 1)`

Factoring out `1/(x^2 + 1)`  yields:

`[f(x) + f(-x)] = (1/(x^2 + 1))*(1/(e^x + 1) + 1/((1/e^x) + 1))`

`[f(x) + f(-x)] = (1/(x^2 + 1))*( 1/e^x + 1 + 1 + e^x)/((e^x + 1)((1/e^x) + 1))`

`` `[f(x) + f(-x)] = (1/(x^2 + 1))*(e^x + 2 + 1/e^x)/(2 + e^x + 1/e^x)`

`` `[f(x) + f(-x)] = (1/(x^2 + 1))`

Integrating `[f(x) + f(-x)]`  yields:

`int_0^1 [f(x) + f(-x)]dx = int_0^1 (dx)/(x^2 + 1)`

`int_0^1 (dx)/(x^2 + 1) = arctan x |_0^1=gt` `int_0^1 (dx)/(x^2 + 1) = arctan 1 - arctan 0 = pi/4`

`` Hence, evaluating the symmetric definite integral yields: `int_0^1 (dx)/(x^2 + 1) = pi/4` .