首先將[1,4]切開為n個(gè)區(qū)間
每個(gè)區(qū)間的底長(zhǎng)Δx = (4 - 1)/n = 3/n
第k個(gè)區(qū)間是[(k - 1)/n,k/n]
選取一點(diǎn)ξ_k = 1 + 3k/n,k∈Z+
所以∫(1→4) f(x) dx
= lim(n→+∞) Σ(k=1→n) f(ξ_k)(Δx_k)
= lim(n→+∞) (3/n) Σ(k=1→n) f(1 + 3k/n)
= lim(n→+∞) (3/n) Σ(k=1→n) [3(1 + 3k/n) + 2]
= lim(n→+∞) (3/n) Σ(k=1→n) (5 + 9k/n)
= lim(n→+∞) (3/n) [5Σ(k=1→n) + 9/n Σ(k=1→n) k]
= lim(n→+∞) (3/n) [5n + (9/n)•n(n + 1)/2]
= lim(n→+∞) (3/n) (19n/2 + 9/2)
= lim(n→+∞) [57/2 + 27/(2n)]
= 57/2