If f : R → R is given by f ( x ) = x + 1 , then the value of lim n → ∞ 1 n f 0 + f 5 n + f 10 n + … . . + f 5 ( n - 1 ) n is:

Step-by-step solution

Let I = lim n → ∞ 1 n f 0 + f 5 n + f 10 n + … . . + f 5 ( n - 1 ) n ⇒ I = ∑ r = 0 n - 1 f 5 r n 1 n ⇒ I = ∫ 0 1 f 5 x d x ⇒ I = ∫ 0 1 5 x + 1 d x ⇒ I = 5 x 2 2 + x 0 1 ⇒ I = 5 2 + 1 = 7 2