You can use sigma notation to write out the Riemann sum for a curve. This is useful when you want to derive the formula for the approximate area under the curve. For example, say that you want to find the approximate area of n right rectangles between x = 0 and x = 3 under the function f (x) = x2 + 1.
By the way, you don’t need sigma notation for the math that follows. It’s just a “convenience” — oh, sure. Cross your fingers and hope your teacher decides not to cover this. It gets pretty gnarly.
The midpoint rule: You can approximate the exact area under a curve between a and b,
with a sum of midpoint rectangles given by the following formula. In general, the more rectangles, the better the estimate.
where n is the number of rectangles,
is the width of each rectangle, x0 through xn are the n + 1 evenly spaced points from a to b, and the function values are the heights of the rectangles.
Applying the midpoint rule to this example, you get:
Here’s the same formula written with sigma notation:
(Note that you could write this instead as
which would more nicely mirror the above formula where the
is on the outside. Either way is fine — they’re equivalent — but you might choose to keep the
on the inside so that the
sum is actually a sum of rectangles. In other words, with the
on the inside, the expression after the
symbol tells you to add up, is the area of each rectangle, namely height times base.)
Now work this out for the six right rectangles in the figure.
You’re figuring the area under x2+ 1 between x = 0 and x = 3 with six rectangles, so the width of each,
Next, because the width of each rectangle is
the right edges of the six rectangles fall on the first six multiples of
These numbers are the x-coordinates of the six points x1 through x6; they can be generated by the expression
where i equals 1 through 6. You can check that this works by plugging 1 in for i in
then 2, then 3, up to 6. So now you can replace the xi in the formula with
The function in this example,
and so now you can write
If you plug 1 into i, then 2, then 3, and so on up to 6 and do the math, you get the sum of the areas of the rectangles in the figure. This sigma notation is just a fancy way of writing the sum of the six rectangles.
Are you having fun? Hold on, it gets worse — sorry. Now you’re going to write out the general sum for an unknown number, n, of right rectangles. The total span of the area in question is 3, right? You divide this span by the number of rectangles to get the width of each rectangle. With 6 rectangles, the width of each is
with n rectangles, the width of each is
And the right edges of the n rectangles are generated by
for i equals 1 through n. That gives you
Or, because f (x) = x2 + 1,
For this last step, you pull the
through the summation symbols — you’re allowed to pull out anything except for a function of i, the so-called index of summation. Also, the second summation in the last step has just a 1 after it and no i. So there’s nowhere to plug in the values of i. This situation may seem a bit weird, but all you do is add up n 1s, which equals n (this is done next).
You’ve now arrived at a critical step. With a sleight of hand, you’re going to turn this Riemann sum into a formula in terms of n.
Now, as almost no one knows, the sum of the first n square numbers,
(By the way, this 6 has nothing to do with the fact that you are using 6 rectangles.) So, you can substitute that expression for the
in the last line of the sigma notation solution, and at the same time substitute n for
The end. Finally! This is the formula for the area of n right rectangles between x = 0 and x = 3 under the function f (x) = x2 + 1.
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