Question

${\int }_{-1}^{1}x(1-x)(1+x)dx$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $1$
• D. $-1$
• E. $0$

Answer 1

The correct option is E $0$

Let $I={\int }_{-1}^{1}x(1-x)(1+x)dx$
Let $f\left(x\right)=x(1-x)(1+x)$
$\Rightarrow f(-x)=(-x)[1-(-x)][1+(-x)]$
$=-x(1+x)(1-x)=-f\left(x\right)$
$\Rightarrow f\left(x\right)$ is an odd function.
Therefore, $I=0$
As, ${\int }_{-a}^{a}f\left(x\right)dx=0$, if $f\left(x\right)$ is an odd function.