Calculate the upper sums Un and lower sums Ln , on a regular partition of the intervals, for the following integrals.

Question

(b)  ∫1 0 (4+18x2)dx

Just want to know which step am I doing wrong

Ln = Σ from i=1 to n of (4 + 18xi^2)Δx

1. Σ from i=1 to n of 4 = 4n
2. Σ from i=1 to n of 18(i/n)^2 = (18/n^2)Σ from i=1 to n of i^2

Σ from i=1 to n of i^2 = (n(n+1)(2n+1))/6

Ln = (1/n) * (4n + (18/n^2) * (n(n+1)(2n+1))/6)

Ln = (1/n) * (4n + (18n(n+1)(2n+1))/6n^2)

Ln = 4 + (18n(n+1)(2n+1))/6n^3)

Answer 1

f(x) = 4 + 18x² is increasing on the interval [0,1]. So, we use left endpoints for L_n. I think that x_i should be (i-1)/n, not i/n. In that way, x₁ = 0, the left endpoint of the first subinterval.