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 the definite integral F(x) and the limit of the Riemann sums, we first need to simplify the expression and understand its relationship to the integral.
Let's start by breaking down the expression:
[(sqrt(36 - ((6/n)^2))) * (6/n)] +
[(sqrt(36 - ((12/n)^2))) * (6/n)] +
... +
[(sqrt(36 - ((6n/n)^2))) * (6/n)]
We can observe that in each term, the value inside the square root, 36 - ((6k/n)^2), can be rewritten as 36 - (36k^2/n^2) by simplifying ((6/n)^2) as 36/n^2.
Now we can simplify further:
[(sqrt((36n^2 - 36k^2)/n^2)) * (6/n)]
By factoring out (6/n) from the square root, we get:
(6/n) * [(sqrt((36n^2 - 36k^2)/n^2))]
Now, let's examine the relationship between this expression and the definite integral F(x) dx from x=0 to 6.
In a right Riemann sum, we approximate the definite integral by partitioning the interval [0, 6] into n subintervals, each of width Δx = (6 - 0)/n = 6/n. The right Riemann sum takes the right endpoint of each subinterval as the representative point.
So, each term in the sum corresponds to one subinterval and represents the area of a rectangle with height (sqrt((36n^2 - 36k^2)/n^2)) and width Δx = 6/n.
The sum of these rectangles approximates the area under the curve F(x) from x=0 to 6, where F(x) is the antiderivative of the integrand.
To find F(x), we integrate the given expression. Since our expression is a right Riemann sum, we integrate each term, [(sqrt((36n^2 - 36k^2)/n^2))] * (6/n), with respect to x from 0 to 6:
∫[(sqrt((36n^2 - 36k^2)/n^2))] * (6/n) dx
Integrating term by term, we get:
(6/n) * ∫[sqrt((36n^2 - 36k^2)/n^2)] dx, evaluated from x=0 to x=6
Integrating sqrt((36n^2 - 36k^2)/n^2), we substitute u = (36k/n) and du = (36/n) dx:
(6/n) * (36/n) * ∫sqrt(1 - u^2) du, evaluated from u=0 to u=6
The definite integral of sqrt(1 - u^2) is sin^(-1)(u), evaluated from u=0 to u=6:
(6/n) * (36/n) * [sin^(-1)(6) - sin^(-1)(0)]
Simplifying further:
(6/n) * (36/n) * (π/2)
(π/2) * [(36/n) * (6/n)]
(3π/2) * (36/n^2)
Now, to find the limit of the Riemann sums as n tends to infinity, we take the limit of this expression as n approaches infinity.
lim (n→∞) [(3π/2) * (36/n^2)]
lim (n→∞) (3π * 36 / (2 * n^2))
As n approaches infinity, the term 36/n^2 approaches zero, and we are left with:
lim (n→∞) (3π/2) = 3π/2
Therefore, the limit of the Riemann sums as n tends to infinity is 3π/2.
To summarize:
- The definite integral F(x) is the antiderivative of each term in the sum, which can be found by integrating [(sqrt((36n^2 - 36k^2)/n^2))] * (6/n) with respect to x from 0 to 6.
- The limit of the Riemann sums as n tends to infinity is 3π/2.