Let the function f : 0 , 2 → ℝ be defined as f x = e min x 2 , x - x , x ∈ [ 0 , 1 ) e x - log e x , x ∈ 1 , 2 , where t denotes the greatest integer less than or equal to t . Then the value of the integral ∫ 0 2 x f x d x is

Step-by-step solution

Given, f x = e min x 2 , x - x , x ∈ [ 0 , 1 ) e x - log e x , x ∈ 1 , 2 ⇒ f x = e x 2 x ∈ [ 0 , 1 ) e , x ∈ 1 , 2 As x - ln x ∈ [ 1 , 2 ) for x ∈ [ 1 , 2 ] , so x - ln x = 1 and x - x = x , so x ∈ 0 , 1 ,   x 2 < x Now solving the integral we get, ∫ 0 2 x f x = ∫ 0 1 x · e x 2 d x + ∫ 1 2 x · e   d x Now in first integral, let x 2 = t ⇒ 2 x d x = d t we get, ⇒ ∫ 0 2 x f x = 1 2 ∫ 0 1 e t   d t + e