If f(x) is an even function, then _ -2 ² (|x| + f(x) x) , d x is

Step-by-step solution

Let I = _ -2 ² (|x| + f(x) x) , dx I = _ -2 ² |x| , dx + _ -2 ² f(x) x , dx Since f(x) is an even function, f(-x) = f(x) . Let g(x) = f(x) x . Then g(-x) = f(-x) (-x) = -f(x) x = -g(x) . Thus, g(x) is an odd function, which gives _ -2 ² f(x) x , dx = 0 . The function |x| is an even function, so _ -2 ² |x| , dx = 2 ₀² x , dx . I = 2 [ x^2 2 ]₀² + 0 I = 4 Answer: 4