Right Riemann sum

Question

The following sum

1/(1+3/n) * 3/n + 1/(1+6/n) * 3/n + 1/(1+9/n) * 3/n + ... + 1/(1+3/n) * 3/n

is a right Riemann sum for a certain definite integral

b

∫ f(x) dx

1

using a partition of the interval [1,b] into n subintervals of equal length.

Then the upper limit of integration must be: b =

and the integrand must be the function f(x) =

Answer 1

Δx = 3/n = (b-a)/n. b is the upper limit of integration and a is the lower limit.

So, since a = 1 and b - a = 3, b must equal 4.

f(x) = 1/x So, the right endpoint of the first interval is 1 + Δx = 1 + 3/n and the height of the first rectangle is f(1+3/n) = 1/(1+3/n). So, the area of the first rectangle is [1/(1+3/n)](3/n). The right endpoint of the second interval is 1 + 2(3/n) = 1+6/n and the height of the second rectangle is f(1+6/n) = 1/(1+6/n). So, the area of the second rectangle is [1/(1+6/n)](3/n), etc.