Let f(x) be a differentiable function on (0, ) such that f'(x) = x e^ 1/x (3x - 1) and f(1) = e . Then the value of _ x x ( f(x) x^3 - 1 ) is equal to

Step-by-step solution

Given f'(x) = x e^ 1/x (3x - 1) . To find f(x) , we integrate f'(x) : f(x) = x e^ 1/x (3x - 1) dx Let t = 1 x x = 1 t dx = - 1 t^2 dt f(x) = 1 t e^t ( 3 t - 1 ) ( - 1 t^2 ) dt f(x) = e^t ( 1 t^3 - 3 t^4 ) dt This is of the form e^t (g(t) + g'(t)) dt = e^t g(t) + C , where g(t) = 1 t^3 and g'(t) = - 3 t^4 . f(x) = e^t t^3 + C = x^3 e^ 1/x + C Now, using f(1) = e : f(1) = 1^3 e^ 1/1 + C = e + C e + C = e C = 0 So, f(x) = x^3 e^ 1/x . We need to evaluate the limit: L = _ x x ( f(x) x^3 - 1 ) L = _ x x ( x^3 e^ 1/x x^3