# Estimating area with finite sums

## Homework Statement

Use the midpoint rule to estimate the area under the graph of f(x) = 7/x and above the graph of f(x) = 0 from [1,25] using two rectangles of equal width.

N/A

## The Attempt at a Solution

So first I found $$\Delta$$x by using (b-a) / n and got (25 - 1) / 2 = 12.

Then I usually proceed by drawing out a number line from 1 to 25 and dividing it up into my 2 segments, each being 12 units wide, i.e. 1---12---25.

Here's where I get stuck. For some odd reason, this midpoint rule throws me off like crazy. I completely understand the left/right endpoint rule, and would have breezed through this problem if it were either of these. This sounds really, really stupid, but I'm not sure how to find the midpoint of each segment without making a mistake. In class we've only gone through examples where each segment is 3 units wide, so the midpoint is very obvious (i.e. 1-3, the midpoint would be 2). How do I do this accurately?

In my mind, I'm thinking that finding the midpoint of each 2 segments would be the same as dividing 1 to 25 into 4 equal parts, so my xi values would be 6 and 18. But when I use these to calculate my area estimate, my answer doesn't match my textbook's. What am I doing wrong, and is there a quick and easy way to find a midpoint that isn't obvious? I feel like I'm over complicating things.

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Well, you found the length of each rectangle (12) but remember it is 12 from 1 not 12 from 0 (hopefully that makes sense).

You would then have 1...13...25. Can you finish from here?

Well, you found the length of each rectangle (12) but remember it is 12 from 1 not 12 from 0 (hopefully that makes sense).

You would then have 1...13...25. Can you finish from here?
Oh, ok. I see what you're saying about the lengths.

So would my midpoints be 7 and 19? What I'm trying to get at is if there's a really easy mathematical way to get the midpoints of these segments. At this point I'm basically being really methodical about it, using my elementary school techniques for finding the median of a set of numbers.

HallsofIvy