Using f(x), determine a formula for the Riemann Sum S_n obtained by dividing the interval [0, 4] into n equal sub-intervals and using the right-hand endpoint for each c_k.
f(x)= 5x+2
Now compute the limit as n goes to infinity S_n to compute the area under the curve over [0,4].
We can start by finding the width of each sub-interval. This can be found by taking the total width of the interval and dividing by n. The interval is from 0 to 4, so the total width is 4.
Δx = 4/n
Next, let's find the right-hand endpoints for each sub-interval. The first right-hand endpoint will be the start of the interval (0) plus the width (Δx). Each subsequent endpoint will be an additional Δx away from the previous endpoint.
c_k = k * (Δx) for k = 1, 2, ..., n
Now, we can substitute c_k into the function f(x) to find f(c_k):
f(c_k) = 5 * (k * (Δx)) + 2
= 5 * (k * (4/n)) + 2
= (20k/n) + 2
Next, we can find the Riemann Sum S_n by summing up the product of f(c_k) and Δx for each value of k:
S_n = Δx * (f(c_1) + f(c_2) + ... + f(c_n))
Now we can substitute in our formulas for f(c_k) and Δx:
S_n = (4/n) * ((20(1)/n + 2) + (20(2)/n + 2) + ... + (20(n)/n + 2))
We can simplify by factoring out the constant terms:
S_n = (4/n) * (20/n + 2 + 40/n + 2 + ... + 20n/n + 2)
S_n = (4/n) * (20(1 + 2 + ... + n)/n + 2n)
Now we can compute the sum of (1 + 2 + ... + n) using the formula for the sum of an arithmetic series:
Sum = (n * (n + 1))/2
So now we have the following:
S_n = (4/n) * (20 * (n * (n + 1))/2n + 2n)
S_n = (4/n) * (10 * (n + 1) + 2n)
Now let's compute the limit as n goes to infinity:
lim (n->infinity) S_n = lim (n->infinity) (4/n) * (10 * (n + 1) + 2n)
To calculate this limit, we can divide each term by n:
lim (n->infinity) S_n = lim (n->infinity) 4 * (10 + 1/n + 2n/n)
As n goes to infinity, the terms with 1/n will approach 0:
lim (n->infinity) S_n = 4 * (10 + 0 + 2)
lim (n->infinity) S_n = 4 * 12
lim (n->infinity) S_n = 48
Thus, the area under the curve over [0, 4] is 48.
To determine a formula for the Riemann Sum, we need to define the width of each sub-interval. In this case, the interval [0, 4] is divided into n equal sub-intervals, so each sub-interval has a width of Δx = 4/n.
Next, we need to determine the right-hand endpoint for each sub-interval, denoted as c_k. For the first sub-interval, c_1 = Δx. For the second sub-interval, c_2 = 2Δx, and so on. Generally, c_k = kΔx.
Now, we can express the Riemann Sum S_n as the sum of the areas of rectangles in the sub-intervals:
S_n = f(c_1)Δx + f(c_2)Δx + f(c_3)Δx + ... + f(c_n)Δx.
Using the given function f(x) = 5x + 2, we substitute the values of c_k and Δx into the formula:
S_n = (5c_1 + 2)Δx + (5c_2 + 2)Δx + (5c_3 + 2)Δx + ... + (5c_n + 2)Δx.
Since c_k = kΔx, we can simplify the formula further:
S_n = (5kΔx + 2)Δx = (5kΔx^2 + 2Δx).
Now, to compute the limit as n goes to infinity (S_n), we substitute Δx = 4/n and simplify further:
S_n = (5k(4/n)^2 + 2(4/n)) = (80k/n + 8/n).
Taking the limit as n goes to infinity, we can see that 80k/n approaches zero while 8/n approaches zero, resulting in:
S_n = 0 + 0 = 0.
Therefore, the limit of the Riemann Sum as n goes to infinity (S_n) is equal to zero, indicating that the area under the curve over the interval [0, 4] is zero.