Calculate ∫[2,0]((𝑥^2)/3)+8𝑑𝑥 by using the definition ∫𝑏𝑎𝑓(𝑥)𝑑𝑥=lim𝑛→∞[∑𝑖=1𝑛𝑓(𝑥𝑖)Δ𝑥] Find Rn and Lim of Rn as n approaches infinity

Question

(1 point) Consider the function 𝑓(𝑥)=((𝑥^2)/3)+8. In this problem you will calculate ∫[2,0]((𝑥^2)/3)+8𝑑𝑥 by using the definition ∫𝑏𝑎𝑓(𝑥)𝑑𝑥=lim𝑛→∞[∑𝑖=1𝑛𝑓(𝑥𝑖)Δ𝑥] The summation inside the brackets is 𝑅𝑛 which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval. Calculate 𝑅𝑛 for 𝑓(𝑥)=𝑥23+8 on the interval [0,2] and write your answer as a function of 𝑛 without any summation signs. You will need the summation formulas in Section 5 of your textbook.

Answer 1

The definition of R_n is the sum from i = 1 to n of f(a + iΔx)Δx, where Δx = (b-a)/n = (2-0)/n = 2/n. Here f(x) = (x^2)/3 + 8 and a = 0, so

f(a + iΔx) = f(2i/n) = (2i/n)^2/3 + 8 = (4i^2)/(3n^2) + 8.

Then R_n is the sum Σ[(4i^2)/(3n^2) + 8]*(2/n) = [8/(3n^3)]*Σ(i^2) + (2/n)*Σ8. Now use the summation formulas

Σ(i^2) = (1/3)n^3 + (1/2)n^2 + (1/6)n and Σ8 = 8n. Then R_n is equal to

[8/(3n^3)]*[(1/3)n^3 + (1/2)n^2 + (1/6)n] + (2/n)*(8n) = (8/9) + 4/(3n) + 4/(9n^2) + 16.

Taking a limit as n goes to infinity gives 8/9 + 16.