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.
idk
To find F(x) and the limit of the Riemann sums as n tends to infinity, we first need to simplify the sum and then evaluate it.
Let's start by simplifying the individual terms in the sum:
[(sqrt(36 - ((6/n)^2))) * (6/n)]
We have a pattern in the sum, where the term varies with n. The term inside the square root has the form 36 - ((6/n)^2), which simplifies to 36 - (36/n^2). Simplifying further, we get 36(1 - 1/n^2).
The term outside the square root is (6/n).
Now, let's rewrite the sum in a simplified form:
[(sqrt(36 - (36/n^2))) * (6/n)] + [(sqrt(36 - (36/n^2 * 4))) * (6/n)] + ... + [(sqrt(36 - (36/n^2 * n^2))) * (6/n)]
Note that each term has 36 - (36/n^2) in common, which cancels out when we factor it out:
[(6/n) * sqrt(36 - (36/n^2))] * [1 + sqrt(36 - (36/n^2 * 4)) + ... + sqrt(36 - (36/n^2 * n^2))]
Now, let's evaluate the sum using sigma notation:
[(6/n) * ∑(k=1 to n) sqrt(36 - (36/n^2 * k^2))]
We have a right Riemann sum for the definite integral F(x) dx from x=0 to 6. In this case, x represents the variable being integrated, and F(x) represents the integrand. To find F(x), we need to find the antiderivative (or integral) of F(x) with respect to x.
Now, let's take the limit as n tends to infinity:
lim(n→∞) [(6/n) * ∑(k=1 to n) sqrt(36 - (36/n^2 * k^2))]
To evaluate this limit, we can use the definition of the definite integral, which is the limit of a Riemann sum as n tends to infinity:
lim(n→∞) [(6/n) * ∑(k=1 to n) sqrt(36 - (36/n^2 * k^2))]
= ∫(0 to 6) F(x) dx
Therefore, the limit of the Riemann sums as n tends to infinity is the definite integral ∫(0 to 6) F(x) dx.
To find F(x), we need to find the antiderivative (integral) of F(x) with respect to x, but without additional information about the function F(x), we cannot determine it.