The following sum
[(sqrt(36-((6/n)^2))).(6/n)] +
[(sqrt(36-((12/n)^2))).(6/n)]+ ... + [(sqrt(36-((6n/n)^2))).(6/n)]
is a right Riemann sum for the definite integral F(x) dx from x=0 to 6
Find F(x) and the limit of these Riemann sums as n tends to infinity.
To find F(x), we need to evaluate the definite integral represented by the Riemann sum.
The given Riemann sum is of the form ∑[(sqrt(36-((6/n)^2))).(6/n)], where n represents the number of subintervals.
Let's break down the integral and calculate it term by term.
We have [(sqrt(36-((6/n)^2))).(6/n)] as one term in the sum. Notice that the form of this term can be rewritten as [(sqrt(36-(36/n^2))).(6/n)].
Now, observe that the expression inside the square root [sqrt(36-(36/n^2))] can be simplified. We factor out 36 from the expression as follows:
sqrt(36(1-(1/n^2)))
Simplifying further:
sqrt(36) * sqrt(1 - (1/n^2))
6√(1 - (1/n^2))
So now we have [(6√(1 - (1/n^2))).(6/n)] for one term in the sum.
Now, let's multiply the terms inside the sum by (6/n) to get:
[(36√(1 - (1/n^2)))/n]
Next, we need to evaluate the definite integral from x=0 to x=6. The integral of F(x) from 0 to 6 is given by:
∫[0 to 6] F(x) dx
Using the limit of Riemann sums, the definite integral can be approximated by taking the limit as n approaches infinity of the Riemann sum.
So, to find F(x), we take the sum of these terms and integrate them from x=0 to x=6.
F(x) = ∫[0 to 6] [(36√(1 - (1/n^2)))/n] dx
Now to find the limit of these Riemann sums as n tends to infinity, we need to simplify the expression inside the integral without the n term.
Let's calculate the integral of [(36√(1 - (1/n^2)))]dx:
∫(36√(1 - (1/n^2))) dx
To integrate this expression, we substitute u = 1 - (1/n^2) and find du.
Differentiating both sides with respect to x, we get:
du/dx = 2(1/n^3)
Rearranging, we have:
dx = n^3/2 du
Substituting these values back into the integral, we get:
∫(36√u) (n^3/2 du)
Simplifying:
(36n^3/2) ∫√u du
Integrating, we have:
(36n^3/2) * (2/3) * u^(3/2) + C
Where C is the constant of integration.
Simplifying further:
(24n^3) * (u^(3/2)/3) + C
Now, substituting back u = 1 - (1/n^2):
F(x) = (24n^3) * ((1 - (1/n^2))^(3/2)/3) + C
We have found the expression for F(x). Now, to find the limit of these Riemann sums as n tends to infinity, we substitute n^3 for n (since we are taking the limit of n^3), and simplify the expression:
lim(n→∞) (24n^3) * ((1 - (1/n^2))^(3/2)/3)
= (24 * ∞^3) * (1^(3/2)/3)
= ∞ * 1/3
= ∞
Therefore, the limit of these Riemann sums as n tends to infinity is infinity.