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) = x^{2} + 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, x_{0} through x_{n} 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,

which 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 x^{2}+ 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 x_{1} through x_{6}; 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 x_{i} in the formula with

giving you

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) = x^{2} + 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) = x^{2} + 1.