If ∫ 0 100 π sin 2 x e x π - x π d x = α π 3 1 + 4 π 2 , α ∈ R where x is the greatest integer less than or equal to x , then the value of α is:

Step-by-step solution

Let, I = ∫ 0 100 π sin 2 x e x π - x π d x We know that if f x is a periodic function, with period T then ∫ 0 n T f x d x = n ∫ 0 T f x d x ,   n ∈ Z ⇒ I = 100 ∫ 0 π sin 2 x e x π - x π d x For, 0 < x < π , 0 < x π < 1 ⇒ x π = 0 ⇒ I = 100 ∫ 0 π sin 2 x e x π d x Now, using cos 2 x = 1 - 2 sin 2 x , we get I = 100 ∫ 0 π e - x / π 1 - cos 2 x 2 d x ⇒ I = 100 × 1