One of the possible functions f(x) which satisfies _ -2 ² f(x) ,dx = 0 is

Step-by-step solution

For an odd function f(x) , the definite integral over a symmetric interval is zero, i.e., _ -a ^ a f(x) ,dx = 0 . Let us check the function in the first option: f(x) = ( 2 - x 2 + x ) Substituting -x for x , we get: f(-x) = ( 2 - (-x) 2 + (-x) ) = ( 2 + x 2 - x ) Using the property of logarithms (a⁻¹) = - (a) : f(-x) = ( ( 2 - x 2 + x )⁻¹ ) = - ( 2 - x 2 + x ) = -f(x) Since f(-x) = -f(x) , the function f(x) = ( 2 - x 2 + x ) is an odd function. Therefore, _ -2 ² ( 2 - x 2 + x ) ,dx = 0 . Answer: ( 2 - x 2 + x )