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## Calculus 2

### Unit 1: Lesson 4

Riemann sums in summation notation

# Worked example: Riemann sums in summation notation

AP.CALC:
LIM‑5 (EU)
,
LIM‑5.B (LO)
,
LIM‑5.B.2 (EK)
Here we express the approximation of the area under a curve in sigma notation. Created by Sal Khan.

## Want to join the conversation?

• How do you determine the accuracy of the estimations you make when using this method?
• Well, an easy way to do that is to compare the area of the Riemann approximation with the area that using the indefinite integral gives you. So to calculate it, it would be R/I, with 'R' being the area given to you through your Riemann approximation, and 'I' being the area that the integral gives for the same area.

For example, suppose we want to check the accuracy of our Riemann approximation for the function x^2 in the section 0-3. We used right-hand rectangles, so we already know this is an over estimation. We decide to use three rectangles in this calculation. That gives us 1+4+9, or 14 un^2. We then integrate the function x^2. That is (x^3)/3, in order the calculate the true area under the section of the curve we take F(3)-F(0), or (3^3)/3-(0^3)/3, which is equal to 3^2, or 9 un^2. We find that we were way off, but how off? 14/9 = 1.555... So we overshot it by about 56 percent.

However, should be mentioned that the accuracy of your approximation is subject to change, because your accuracy will vary depending on how many rectangles you make, and how the function itself behaves. You see, in the function x^2 our approximations will get less and less accurate as we go up the x-axis because the curve is getting steeper and steeper, whereas in the function f(x) = ln(x) we would find that we get more accurate as we go up the x-axis because the rate of change is getting smaller and smaller (approaching zero, in fact).
• why do we need to use riemann's approximation when we can just use the indefinite integral?
• Because the indefinite integral doesn't tell you anything about the area, its just another function. But if you are referring to the definite integral (which is basically just the indefinite integral evaluated at two endpoints), the reason we have to learn Riemann Sums is that this technique is actually the rigorous foundation of the definite integral. The definite integral is more or less defined as the limit of a Riemann Sum, taking the limit as the number of rectangles goes to infinity (or equivalently, as the width of each rectangle goes to zero).

So, its important to understand where the definite integral comes from. In the same way that its important to realize that the true (if you will) definition of the derivative is the limit as h goes to zero of (f(x+h)-f(x))/h, its important to realize that the whole concept of the definite integral is valid because its defined, at its very foundation, as a limiting case of a Riemann sum.

Hope that helps.
• Hey, I'm not quite sure if I understand where you got the formula f(2n-1) from. I do understand that that's the height of any rectangle. But how did u find this function from just looking at it?

Best regards
• In the video, he says that it seems like the heights of the rectangles are based on the midpoints of the rectangles. The midpoints of the rectangles are located at the odd numbers `1, 3, 5, 7`. A general formula for these odd numbers is `2n - 1`, where `n` goes from `1` to `4` (inclusive). Therefore, the height of the `n`th rectangle is `ƒ(2n - 1)`.

If `2n - 1` still seems mysterious, try plugging in some values for `n`; it will become apparent that you get odd integers. The ability to recognise such formulas comes with experience.
• in the sigma notation where does he get the value n=1?
• We are just counting rectangles and counting begins at 1. In general, In general, there are, there are n rectangles. In this case there are 4, so we sum from the first, that is, n = 1 to the last, that is, n=4. The first rectangle is n=1, the second n=2 etc. It is just a coincidence that the center of the first rectangle also happens to be centered on the point x=1.
Hope that helped!
• Can you just put the two out in front of the sigma? I mean, Sal kind of ends up doing that anyways...
• Yes, absolutely. You're just factoring it out of each term in the sum.
• Um, there was a mistake in the video. He put a times sign when he needed to use a plus sign between f(5) and 2*f(7)
• correct, you can put it on tips & thanks section so KA could see it. Good eyes!!
• It is true that more the number of rectangles we choose, better is the accuracy we get. However, this curve has something interesting. It appears that accuracy is also achieved, rather better, by choosing less number of rectangles (thus having wider rectangles) and height as the mid boundary (At ) of rectangle, because overestimating the area at the left half of rectangle is compensated by underestimating the area above the right half of that rectangle. Can someone explain to me that this advantage is specific to these types of curves and does not work in general.
• Are you talking about the fact that it looks like the part of the rectangle that protrudes above the curve on the left seems to fill the piece missing underneath the curve? I was also wondering about this. If true, then wouldn't taking the area of the rectangle work?
• Why is Sigma being used what is the significance of sigma?
• It is common practice in mathematics, for some reason, to use greek letters to denote various quantities. A capital sigma, `∑`, happens to be used to denote sums and series. There is no real significance of the symbol `∑`; you could come up with your own notation for sums, something very different, but it would be annoying if everyone did so. Having a convention is nice. `∑` happens to be the convention.
• Would this be an underestimate or an overestimate?