# Question #bd150

Nov 18, 2016

The integral is equal to $0$.

#### Explanation:

Notice that $f \left(x\right) = {x}^{5} - 6 {x}^{9} + \sin \frac{x}{1 + {x}^{4}} ^ 2$ is an odd function, that is, $f \left(- x\right) = - f \left(x\right)$.

Odd functions have rotational symmetry about the origin, which means that whatever is in the positive $x$ region of the graph is its opposite on the opposite side of the graph.

This should be clear looking at another odd function, like ${x}^{5} - 4 {x}^{3}$:

graph{x^5-4x^3 [-14.24, 14.24, -7.12, 7.12]}

We can use this property of odd functions to surmise the rule that when $f \left(x\right)$ is an odd function, ${\int}_{- a}^{a} f \left(x\right) \mathrm{dx} = 0$, since the positive areas on one side will cancel with the corresponding negative areas on the other.

This is the case with the given integral, so we can quickly conclude that ${\int}_{- 1}^{1} \left({x}^{5} - 6 {x}^{9} + \sin \frac{x}{1 + {x}^{4}} ^ 2\right) \mathrm{dx} = 0$.