If f(x) is defined in [-2,2] by f(x)=4 x²-3 x+1 and g(x)= f(-x)-f(x) (x²+3 ) , then _ -2 ² g(x) d x=

Step-by-step solution

Given that, f(x)=4 x²-3 x+1g(x)= f(-x)-f(x) x²+3 Therefore, 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. Thus, _ -2 ² g(x) d x=0