# How would show int_(-1)^1 f(x)dx =0 without calculate integral? f(x)=arctgx

You need to remember the basic property of an odd function `f(x)` , such that:

`int_(-a)^a f(x)dx = 0`

Since you need to prove that `int_(-1)^1 f(x)dx = 0` , you should prove that `f(x)` is an odd function, such that:

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

You need to replace -`x` for...

## Unlock This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

You need to remember the basic property of an odd function `f(x)` , such that:

`int_(-a)^a f(x)dx = 0`

Since you need to prove that `int_(-1)^1 f(x)dx = 0` , you should prove that `f(x)` is an odd function, such that:

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

You need to replace -`x` for `x` in equation of the function, such that:

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

You need to remember that `arctan(-theta) = -arctan theta` , hence, reasoning by analogy, yields:

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

Since `f(-x) = -f(x)` , the given function `f(x) = arctan x` is an odd function and, by definition, `int_(-a)^a f(x) dx = 0` .

Hence, testing if `int_(-1)^1 arctan x dx= 0` , using the properties of an odd function, yields that the statement `int_(-1)^1 arctan x dx= 0` holds.