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 first notice that each summand in the given expression is of the form [sqrt(36 - ((6k/n)^2)) * (6/n)]. Let's simplify this expression:
sqrt(36 - ((6k/n)^2)) * (6/n)
= sqrt(36 - (36 * k^2 / n^2)) * (6/n)
= sqrt(36 * (1 - (k^2 / n^2))) * (6/n)
= (6/n) * sqrt((6^2 * (1 - (k^2 / n^2)))).
Now, let's rewrite the given sum using this simplified expression:
[(sqrt(36 - ((6/n)^2))).(6/n)] + [(sqrt(36 - ((12/n)^2))).(6/n)] + ... + [(sqrt(36 - ((6n/n)^2))).(6/n)]
= (6/n) * sqrt((6^2 * (1 - (1/n^2)))) + (6/n) * sqrt((6^2 * (1 - (2^2/n^2)))) + ... + (6/n) * sqrt((6^2 * (1 - (n^2/n^2))))
= (6/n) * [sqrt(6^2 * (1 - (1/n^2))) + sqrt(6^2 * (1 - (2^2/n^2))) + ... + sqrt(6^2 * (1 - (n^2/n^2)))].
Now, let's consider the limit of this expression as n tends to infinity. Notice that the sum inside the bracket is a Riemann sum for the definite integral of sqrt(6^2 * (1 - x^2)) over the interval [0, 1]. Thus, this Riemann sum approaches the definite integral as the partition gets finer (or as n tends to infinity).
Therefore, the limit of the given Riemann sum as n tends to infinity is equal to the definite integral:
F(x) = ∫[0,6] sqrt(6^2 * (1 - x^2)) dx.
To evaluate this integral, we can make use of the trigonometric substitution x = 6sin(theta):
dx = 6cos(theta) d(theta),
and the limits of integration become [0, pi/2] since sin(theta) ranges from 0 to 1 when theta varies from 0 to pi/2.
Now, substituting and simplifying:
F(x) = ∫[0,pi/2] sqrt(6^2 * (1 - (6sin(theta))^2)) (6cos(theta)) d(theta)
= 36 ∫[0,pi/2] sqrt(36 - 36sin^2(theta)) cos(theta) d(theta)
= 36 ∫[0,pi/2] sqrt(36cos^2(theta)) cos(theta) d(theta)
= 36 ∫[0,pi/2] |6cos(theta)| cos(theta) d(theta)
= 36 ∫[0,pi/2] 6cos^2(theta) d(theta).
Using the identity cos^2(theta) = (1 + cos(2theta))/2, we have:
F(x) = 36 ∫[0,pi/2] 6(1 + cos(2theta))/2 d(theta)
= 18 ∫[0,pi/2] (1 + cos(2theta)) d(theta)
= 18 [theta + (sin(2theta))/2] |[0,pi/2]
= 18 [(pi/2 + (sin(pi))/2) - (0 + (sin(0))/2)]
= 18 [(pi/2 + 0) - (0 + 0)]
= 18 * (pi/2)
= 9pi.
Therefore, F(x) = 9pi, and the limit of the given Riemann sums as n tends to infinity is equal to 9pi.
To find F(x), we need to evaluate the definite integral of the function F(x) over the interval from x=0 to x=6. Let's calculate it step by step.
First, let's express the given sum as a general form:
[(sqrt(36 - ((6/n)^2))) * (6/n)] + [(sqrt(36 - ((12/n)^2))) * (6/n)] + ... + [(sqrt(36 - ((6n/n)^2))) * (6/n)]
Now, notice that each term in the sum has the same pattern: sqrt(36 - ((6k/n)^2)) * (6/n), where k is the index of each term ranging from 1 to n.
To simplify the expression, let's focus on one of the terms:
sqrt(36 - ((6k/n)^2)) * (6/n)
Using some algebraic manipulation, we can rewrite this term as:
(6 * sqrt(36 - (36k^2/n^2)))/n
Now, let's take a closer look at this expression:
- The numerator, 6 * sqrt(36 - (36k^2/n^2)), represents the height of each rectangle in the Riemann sum.
- The denominator, n, represents the width of each rectangle in the Riemann sum.
Therefore, we can see that each term in the sum represents the area of a rectangle with height (6 * sqrt(36 - (36k^2/n^2))) and width (6/n).
To find the definite integral, F(x), we need to take the limit as n tends to infinity to ensure that we have an infinite number of rectangles that approximates the area perfectly.
To find F(x), we will evaluate the limit of this Riemann sum as n tends to infinity:
lim(n→∞) [(sqrt(36 - ((6/n)^2))) * (6/n)] + [(sqrt(36 - ((12/n)^2))) * (6/n)] + ... + [(sqrt(36 - ((6n/n)^2))) * (6/n)]
Now, let's simplify this limit step by step. Recall that for the definite integral, we want to compute the area under the curve from x=0 to x=6, so our limit expression can be written as:
lim(n→∞) Σ[(6 * sqrt(36 - (36k^2/n^2)))/n], where k ranges from 1 to n.
Next, we can simplify the summation notation by expressing it as a definite integral:
lim(n→∞) Σ[(6 * sqrt(36 - (36k^2/n^2)))/n] = lim(n→∞) ∫[f(x) dx] from x=0 to x=6
Now, let's determine the function f(x). Note that in the given sum, x is equal to (6k/n), so we can write it as:
f(x) = 6 * sqrt(36 - 36(x/6)^2)
Finally, we have:
F(x) = ∫[f(x) dx] from x=0 to x=6
To find the exact value of F(x), you will need to evaluate this definite integral using techniques such as substitution or integration by parts.
As for the limit of the Riemann sum as n tends to infinity, it represents the exact value of the definite integral F(x). Therefore, the limit as n tends to infinity is equal to F(x).