If f(x) is defined in [-2, 2] by f(x) = 4x2 - 3x +1 and g(x)=(f(-x)-f(x)/x2+3) then ∫-22 g(x)dx =

Question

If f(x) is defined in [-2, 2] by f(x) = 4x2 - 3x +1 and g(x)=(f(-x)-f(x)/x2+3) then ∫-22 g(x)dx =  (A) 24 (B) 0 (C) -48 (D) 64

Tardigrade Admin · Accepted Answer

Given that f(x) =4 x2-3 x+1, g(x)=(f(-x)-f(x)/x2+3) ∴ g(x) =((4 x2+3 x+1)-(4 x2-3 x+1)/x2+3) =(6 x/x2+3) Now, g(-x) =-(6 x/x2+3) =-g(x) Which is an odd function ∫ limits-22 g(x) d x=0

Q. If $f (x)$ is defined in $[- 2, 2]$ by $f (x) = 4 x^{2} - 3 x + 1$ and $g (x) = \frac{f ( - x ) - f ( x )}{x ^{2} + 3}$ then $\int_{- 2}^{2} g (x) d x =$

24

0

-48

64

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

Q. If f(x) is defined in [−2,2] by f(x)=4x2−3x+1 and g(x)=x2+3f(−x)−f(x)​ then ∫−22​g(x)dx=

24

0

-48

64

Given that f(x)=4x2−3x+1,g(x)=x2+3f(−x)−f(x)​ ∴g(x)=x2+3(4x2+3x+1)−(4x2−3x+1)​ =x2+36x​ Now, g(−x)=−x2+36x​ =−g(x) Which is an odd function −2∫2​g(x)dx=0

Q. If $f (x)$ is defined in $[- 2, 2]$ by $f (x) = 4 x^{2} - 3 x + 1$ and $g (x) = \frac{f ( - x ) - f ( x )}{x ^{2} + 3}$ then $\int_{- 2}^{2} g (x) d x =$