Question

Let $g\left(x\right)=\frac{1}{2}\left[f\right(x)-f(-x\left)\right]$ for $-3\le x\le 3$ and $f\left(x\right)=2x^2-2x+1$ then find ${\int }_{-3}^{3}g\left(x\right)dx$.

Answer 1

Given, $f\left(x\right)=2x^2-2x+1$.

Then

$g\left(x\right)=\frac{1}{2}\left[f\right(x)-f(-x\left)\right]$

$=-2x$.

So $g\left(x\right)$ is and odd function as $g(-x)=-g\left(x\right)$.

$\therefore {\int }_{-3}^{3}g\left(x\right)dx=0$ [ Due to oddness of $g\left(x\right)$]