How do you prove without calculation `int_-a^a (x^2*sin2x dx)/(x^2+1)`  = 0?

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to remember how to exploit the symmetry of a function when evaluating the definite integral of this function.

- if f(x) is odd, then `int_-a^a f(x) dx = 0`

Since the problem claims that the definite integral of symmetric function is cancelling, then f(x) must be an odd function.

You need to remember the behaviour of an odd function:

`f(-x) = -f(x)`

You need to plug -x in the equation of the given function such that:

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

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

This proves that the function `f(x) = (x^2*(sin 2x))/(x^2+1)`  is odd, hence `int_-a^a f(x) dx = 0` .

We’ve answered 318,994 questions. We can answer yours, too.

Ask a question