Question

If $f\left(x\right)$ is defined $[-2,2]$ by $f\left(x\right)=4x^2-3x+1$ and $g\left(x\right)=\frac{f(-x)-f\left(x\right)}{x^2+3}$, then ${\int }_{-2}^{2}g\left(x\right)dx$ is equal to

• A. $64$
• B. $-48$
• C. $0$
• D. $24$

Answer 1

The correct option is C $0$

Given that
$f\left(x\right)=4x^2-3x+1,$ $g\left(x\right)=\frac{f(-x)-f\left(x\right)}{x^2+3}$
Therefore, $g\left(x\right)=\frac{(4x^2+3x+1)-(4x^2-3x+1)}{x^2+3}$
$=\frac{6x}{x^2+3}$
Now, $g(-x)=-\frac{6x}{x^2+3}=-g\left(x\right)$
Which is an odd function
Thus ${\int }_{-2}^{2}g\left(x\right)dx=0$