Question

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

• A. 64
• B. $-48$
• C. 0
• D. 24

Answer 1

Answer: (C)

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