Find the limit of these Reimann sums

Question

for the following sum 

√(36-(6/n)²)(6/n) +√(36-(12/n)²)(6/n)+…+√(36-(6n/n)²)(6/n)

is a right Reimann sum for the definite integral ∫f(x)dx with upper bound b and lower bound 0 

what is b?

what is f(x)?

what is the limit of these Reimann sums as n approaches infinity? 

Answer 1

b = 6. Simplify the (6n/n) bit inside the last square root.

f(x) = √(36 - x^2). Replace any given parentheses with a square by x^2.

As written, f(x) denotes the top half of a circle with radius 6 centered at the origin.

The area of the full circle of this radius would be 36π, so the semicircle has area 18π.

Integrating from 0 to b = 6 calculates the area in only the right half of the semicircle defining f(x)

This means the limit of the Riemann sum as n approaches infinity is 9π.