The following Summation
1 2 1 2 1 2
------ x - + ------ x - + ------ x -
1+(2/n) n 1+(4/n) n 1+(6/n) n
is a right Riemann sum for this integral
�ç f(x)dx With subintervals [1,3]
can someone help me solve for the inegrand which is f(x).?
To solve for the integrand, f(x), we need to find the function that corresponds to the given right Riemann sum.
First, let's break down the given right Riemann sum:
(1/(1 + (2/n))) * n/x - (1/(1 + (4/n))) * n/x + (1/(1 + (6/n))) * n/x
We can simplify this expression further by removing the common factors of n/x from each term:
(1/(1 + (2/n))) - (1/(1 + (4/n))) + (1/(1 + (6/n)))
Simplifying the denominators, we have:
(1/(1n/n + 2)) - (1/(1n/n + 4)) + (1/(1n/n + 6))
Now, let's find the limit of this expression as n approaches infinity, which will give us the integrand, f(x).
As n approaches infinity, the terms with n in the denominators become negligible compared to 1. Therefore, we can simplify the expression further:
(1/2) - (1/4) + (1/6)
Finding the common denominator, we have:
(3/6) - (1/6) + (2/6)
Combining the terms, we get:
4/6
Simplifying this fraction, we have:
2/3
Therefore, the integrand, f(x), is equal to 2/3.