Use a Riemann sum with n = 3 terms and the right endpoint rule to approx. ∫(1, 2) sin(1/x)dx.

My teacher just needs the terms written out, no need to add or multiply. This is a problem she did up on the board, so here's her answer:

sin(4/3)(1/3) + sin(5/3)(1/3) + sin(2)(1/3)

I might have misread her writing, but is this right? Wouldn't the (1/x) in sin(1/x) create:

sin(3/4)(1/3) + sin(3/5)(1/3) + sin(1/2)(1/3)

well, just looking at the rightmost term where x = 2

we have sin (1/2)
so I believe that you are correct and the teacher wrote sin( x ) instead of sin (1/x).

To approximate the integral ∫(1, 2) sin(1/x)dx using a Riemann sum with n = 3 terms and the right endpoint rule, you need to partition the interval [1, 2] into three subintervals of equal width.

First, calculate the width of each subinterval:
Δx = (2 - 1) / 3 = 1/3

Next, identify the right endpoints of each subinterval:
x1 = 1 + Δx = 1 + (1/3) = 4/3
x2 = 1 + 2Δx = 1 + 2(1/3) = 5/3
x3 = 1 + 3Δx = 1 + 3(1/3) = 2

Now, let's calculate the approximate integral using the right endpoint rule:

For the first subinterval [1, 4/3]:
Approximation for the integral on this subinterval:
(sin(1/4/3))(1/3)

For the second subinterval [4/3, 5/3]:
Approximation for the integral on this subinterval:
(sin(1/5/3))(1/3)

For the third subinterval [5/3, 2]:
Approximation for the integral on this subinterval:
(sin(1/2))(1/3)

Therefore, the Riemann sum approximation of the integral becomes:
(sin(4/3))(1/3) + (sin(5/3))(1/3) + (sin(1/2))(1/3)

So, based on your teacher's answer, it seems like you have correctly rewritten the terms. The correct expression is:
(sin(4/3))(1/3) + (sin(5/3))(1/3) + (sin(1/2))(1/3).