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.
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].
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