Let f: R R be a twice differentiable function satisfying f''(x) + f''( - x) = 0 for all x R . If f(0) = 2 and f( ) = 6 , then the value of the integral ₀^ f(x) x dx is equal to

Step-by-step solution

Given f''(x) + f''( - x) = 0 . Integrating both sides with respect to x : f''(x) dx + f''( - x) dx = C f'(x) - f'( - x) = C Substituting x = 2 : f' ( 2 ) - f' ( - 2 ) = C f' ( 2 ) - f' ( 2 ) = C C = 0 Thus, f'(x) - f'( - x) = 0 . Integrating again with respect to x : f'(x) dx - f'( - x) dx = K f(x) + f( - x) = K Substituting x = 0 : f(0) + f( ) = K Given f(0) = 2 and f( ) = 6 , we get K = 2 + 6 = 8 . Therefore, f(x) + f( - x) = 8 for all x R . Let I = ₀^ f(x) x dx (1) Using the property ₀^a g(x) dx = ₀^a g(a-x) dx