Consider the function f(x)=((x^2)/3)-6. In this problem you will calculate the integral from 0 to 4 by using the definition lim n approches infinity sigma...

Question

 Consider the function f(x)=((x^2)/3)-6.

In this problem you will calculate the integral from 0 to 4 by using the definition lim n approches infinity sigma...

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)=((x^2)/3)-6 on the interval [0,4]  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. 

Rn=???

lim n approaches infnity Rn=???

Answer 1

As we are evaluating from 0 to 4, x will be evaluated at 4i/n for i = 1 to n. f(x_i) = (4i/n)²/3 - 6

∑₁ⁿ 4/n ((4i/n)²/3 - 6) =

4/n ((16/(3n²)n(n+1)(2n+1)/6 - 6n) = 64/9 + 32/(3n)+ 32/(9n²)- 24= -152/9 + 32/(3n) + 32/(9n²)

The limit as n goes to infinity is -152/9

Answer 2

Δx = (4 - 0)/n = 4/n.

Then R_n = ∑_i=1ⁿ((4/ni)²/3 - 6)·4/n = ∑_i=1ⁿ(16/(3n²)i² - 6) = 64/(3n^·3)n(n + 1)(2n+ 1)/6 - 6n·4/n =

32(n + 1)(2n + 1)/(9n²) - 24;

lim R_n as n → ∞ = 64/9 - 24 = - 152/9