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

Question

Dr. Abubacker · Accepted Answer

(c)
Given, $f(x) = 4x^{2} - 3x + 1,g(x) = \frac{f( - x) - f(x)}{x^{2} + 3}$
$	herefore g(x) = \frac{\left( 4x^{2} + 3x + 1 ight) - (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
$	herefore\int_{- 2}^{2}{g(x)dx = 0}$

Q. If $f(x)$ is defined $\lbrack - 2,\ 2\rbrack\ by\ f(x) = 4x^{2} - 3x + 1$ and $g(x) = \frac{f( - x) - f(x)}{x^{2} + 3}$, then $\int_{- 2}^{2}{g\ (x)dx}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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