Find f(x) based on the following Riemann sum

Question

The following sum

√(4-(2/n)²)*(2/n) + √(4-(4/n)²)*(2/n) + ... + √(4-(2n/n)²)*(2/n)

is a right Riemann sum with n subintervals of equal length for the definite integral

2

∫ f(x) dx

0

where f(x) = ???

Answer 1

First, put your sum in sigma notation:

∑ⁿ_i=1 (4 - (2i / n)²)^1/2 (2 / n)

The definition of definite integral in sigma notation and limit is:

∫^b_a f(x)dx = lim_n→∞ ∑ⁿ_i=1 f(a + Δx•i) Δx

where: Δx = (b-a)/n

Looking at the sigma notation of your sum:

Δx = (b-a)/n = 2 / n

∴ b-a = 2

b = 2 and a = 0 to verify your integral. So if we compare the sigma notation of your sum and the definition of integral, we can have the following:

f(a + Δx•i) = (4 - (2i / n)²)^1/2

Plug in the value of a and Δx:

f(0 + 2i / n) = (4 - (2i / n)²)^1/2

f( 2i / n) = (4 - (2i / n)²)^1/2

∴ f(x) = (4 - x²)^1/2

Answer 2

f(x) = √(4 - x²)