# What is integral of function f(x) in interval [1;2] when f(x)=x^3+2*f(1/x)?

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### 1 Answer

Consider the given relation and substitute x by 1/x.

`f(1/x) = (1/x^3)+ 2*f(x)`

Consider f(x) the unknown in the system of equations:

`f(1/x) = (1/x^3)+ 2*f(x)`

`f(x) = x^3+ 2*f(1/x)`

Substitute the relation expressing `f(1/x)` in the first equation.

`f(x) = x^3 + 2*[(1/x^3)+ 2*f(x)]`

`f(x) = x^3 + 2/x^3 + 4*f(x)`

`x^3 + 2/x^3 + 4*f(x) - f(x) = 0`

`x^3 + 2/x^3 + 3*f(x) = 0`

`f(x) = -(x^3 + 2/x^3)/3`

Calculate definite integral of function f(x).

`int_1^2f(x)dx` = -`int_1^2(x^3/3)dx` -`int_1^2(2x^-3/3)dx`

`int_1^2f(x)dx` =`1/12 - 8/12+ (2/3)*(1/8 + 1/2)`

`` `int_1^2f(x)dx` = `|-7/12 + 5/12| = 1/6`

**ANSWER: The integral of function f(x)=1/6**