Odd?

Calculus Level 1

$\large \int _{ -1 }^{ 1 }{ \frac { \sin { x } }{ 1+{ x }^{ 2 }+{ x }^{ 4 } } \, dx } = \, ?$

0

3

\pi

4

2 solutions

Chew-Seong Cheong
Apr 19, 2016

The title of the problem gives away the solution. The function $f(x) =\dfrac {\sin (x)} {1+x^2 +x^4}$ is an odd function. Therefore, $f(-x) =-f(x)$ .

$\begin{aligned} \int_{-1}^1 f(x) \, dx & = \int_0^1 f(x) \, dx + \int_{-1}^0 f(x) \, dx \\ & = \int_0^1 f(x) \, dx - \int_0^1 f(x) \, dx \\ & = \boxed {0}\end{aligned}$

I understand how an odd function over an odd function is 0, but what determines an odd function?

massimo 22 - 3 years, 6 months ago

an odd function is basically like this sin(-x)=-sin(x) so when u substitute -x in places of x like substituting in the function f(x) as f(-x) then u will see that the answer will be negative of the original like f(-x)=-f(x) which satisfies the requirement for odd function!!!

erica phillips - 3 years, 3 months ago

f(x) is an odd function so its primitive is also odd that's all :)

Sïçk ßøy - 2 years, 4 months ago

Odd function so f(x) = -f(x) so the limits 1 and -1 cancel each other out so it is 0.

Joseph LAU [11D] - 10 months ago

Aneeq Designs
Jul 12, 2019

function is odd

Odd?

2 solutions

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