Why is the Riemann Sum 1 + 6i/n and Not 1 + 3i/n?

[Houdini1]

Question:
A solid has a rectangular base that lies in the first quadrant and is bounded by the x and y-axes and the lines x=2, y=1. The height of the solid above point (x,y) is 1+3x. Find the Riemann approximation of the solid.

Solution:

I already know that the solution is \(\displaystyle \sum_{i=1}^{n} \frac{2}{n} \left(1+\frac{6i}{n} \right)\). What I don't see is why it's 1+(6i)/n and not 1+(3i)/n. Volume can be generalized to be the area of the base times the height, so for this problem I have something like \(\displaystyle x*y*f(x)\). Of course x is changing so I must rewrite this.

For any partition where we approximate the volume between x_1 and x_2 the length will be \(\displaystyle \Delta x=\frac{2}{n}\) The y value is a constant 1, so won't need to be written explicitly as far as I can see. I know this part is incorrect but it seems to me that the height should be \(\displaystyle 1+\frac{3i}{n}\), but I know that since we haven't defined where \(\displaystyle x_i\) is in each partition (it could be the left value, middle value, right value or anywhere) then I'm really stuck here.

EDIT: Now that I think about it more since we haven't defined what n is we don't know what i/n either and i/n will change according to how many partitions we take. So am I correct in thinking that \(\displaystyle f(x_i)=f(i\Delta x)\)? If so this makes sense now.

 

[CaptainBlack]

Question:
A solid has a rectangular base that lies in the first quadrant and is bounded by the x and y-axes and the lines x=2, y=1. The height of the solid above point (x,y) is 1+3x. Find the Riemann approximation of the solid.

Solution:

I already know that the solution is \(\displaystyle \sum_{i=1}^{n} \frac{2}{n} \left(1+\frac{6i}{n} \right)\). What I don't see is why it's 1+(6i)/n and not 1+(3i)/n. Volume can be generalized to be the area of the base times the height, so for this problem I have something like \(\displaystyle x*y*f(x)\). Of course x is changing so I must rewrite this.

For any partition where we approximate the volume between x_1 and x_2 the length will be \(\displaystyle \Delta x=\frac{2}{n}\) The height is a constant 1, so won't need to be written explicitly as far as I can see. I know this part is incorrect but it seems to me that the height should be \(\displaystyle 1+\frac{3i}{n}\), but I know that since we haven't defined where \(\displaystyle x_i\) is in each partition (it could be the left value, middle value, right value or anywhere) then I'm really stuck here.

EDIT: Now that I think about it more since we haven't defined what n is we don't know what i/n either and i/n will change according to how many partitions we take. So am I correct in thinking that \(\displaystyle f(x_i)=f(i\Delta x)\)? If so this makes sense now.

This is just a one dimensional integral/Riemann sum problem in disguise, since the height is a function of \(x\) only and the interval for \(y\) is of width \(1\), so we seek:
\[\int_{x=0}^2 (1+3x) \; dx\]
We divide the \(x\)-interval \([0,2]\) into \(n\) strips each of width \(2/n\), then the area of the \(i\)-th strip is approximately:
\[A_i=\frac{2}{n} \times \left(1+3 \times \frac{2\times i}{n}\right)\]
where we are using the \(x\) value of the right edge of the \(i\)-th strip (\( (2 \times i)/n \) ) as our approximate value of \(x\) in the formula for the height of the for the strip.

Then the right Riemann sum is:
\[S_n=\sum_{i=1}^{n}A_i={\Large{\sum_{i=0}^{n-1}}}\left[ \frac{2}{n} \times \left(1+3 \times \frac{2\times i}{n}\right)\right] \].

CB

 

[Houdini1]

Almost makes perfect sense, just would like clarification about the \(\displaystyle f(x_i)=f(i \times \Delta x)\) part. If that's true than I see \(\displaystyle f(x_i)\) as \(\displaystyle \left( 1+ 3 \left[ i \times \frac{2}{n} \right] \right)\). I know that's just rewriting what you posted but writing it this way seems to confirm my stated assumption.

That's well put that this is a disguised single variable integral too.