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 :

Solution

Maths sir · 4 steps

Put 5r/n into f first, and the sum turns into ∫₀¹ (5x+1) dx.

1. Write the general term. The terms are f(5r/n) for r = 0, 1, …, n-1. Since f(x) = x + 1, each term is 5r/n + 1.
2. Spot the Riemann sum. So the expression is 1/nΣᵣ₌₀ⁿ⁻¹ g(r/n) with g(x) = 5x + 1. As n → ∞, r/n becomes x, 1/n becomes dx, and r/n runs from 0 to 1.

   lim(n → ∞) 1/nΣᵣ₌₀ⁿ⁻¹(5r/n+1) = ∫₀¹ (5x+1) dx = [5x²/2+x]₀¹ = 5/2+1

3. Check directly. You can also add the terms directly: 1/n[n + 5/n × n(n-1)/2] = 1 + 5(n-1)/2n. This tends to 1 + 5/2 as n → ∞.
4. So. The limit is 7/2, which is option (d).

The idea to keep. For a limit of a sum, write the r-th term as 1/n g(r/n). Then replace r/n by x and 1/n by dx. A scale factor like the 5 stays inside g.

Answer from the NTA final answer key. Solution by PrepWiser.