Main content

### Course: Integral Calculus > Unit 1

Lesson 4: Riemann sums in summation notation- Riemann sums in summation notation
- Riemann sums in summation notation
- Worked example: Riemann sums in summation notation
- Riemann sums in summation notation
- Midpoint and trapezoidal sums in summation notation
- Riemann sums in summation notation: challenge problem

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Riemann sums in summation notation: challenge problem

When a function is negative, Riemann sums seem to treat it as having "negative area".

## Want to join the conversation?

- Confused me as I thought sum might still mean the total of a positive and negative value(53 votes)
- Although it sounds tempting, area cannot be negative. When a function dips below the x-axis, the space between the function and the x-axis should be treated the same way as if it were above the x-axis. Because there is no absolute value sign, both II and III are wrong as they subtract area from themselves.(47 votes)

- So, can you actually use the Riemann sum at all to work out the area of the rectangles that are in the third/fourth quadrants? Or is it just impossible because you'd never be able to get a positive number?(11 votes)
- You could always introduce the absolute value function into the sum. To say this more precisely, instead of doing the summation of that function, you can do the summation of the absolute value of that function. In a sense, this would force all the values of the function to be positive which makes it give you the height of every rectangle.(22 votes)

- It confused me when Sal ruled out the second choice. Traditionally, you cannot have a negative area, but I thought that area under the x-axis above the curve was considered negative. At least, it is when evaluating signed area.(12 votes)
- The way that areas below the x-axis are interpreted when integrating depends on what the function is representing. In other words, is the function representing something such that any area below the x-axis should be subtracted or added to the area above the x-axis. So, there is not a single answer -- what you do depends on what the area is representing.(7 votes)

- For the part where you discuss whats inside the f(x), how is that related to whether it is negative or positive? its unclear why the statements are true and false(7 votes)
- I was confused about this as well. You have to remember that the f(stuff) is directing you to plug the value of the "stuff" you just found into whatever function you're dealing with. For example, the first red rectangle you would have f( -1 + 1/2 ) which equals f( -1/2 ). When you plug -1/2 into the equation being graphed you get some very small negative value right below the x-axis. That small negative value is what causes the trouble since no matter what red rectangle you're referencing, the corresponding f(stuff) = y = "rectangle height" value is always going to be negative which will give you a negative approximate area. TL;DR a rectangle below the x-axis may be 4 units tall, but the y-value you're using to determine that height is actually negative and that y-value is what is being used in the sum, not the absolute value of y which is the height.(6 votes)

- confuses me why we dont say what we said for second (it can not be negative) also for first one(6 votes)
- We don't take the x-values into consideration when calculating the area, just the distance between them (1/2) which is always positive so it doesn't matter whether the x-values are positive or negative.

We do, however, use the y-values as the height which can be positive or negative and that's where we get into the issue of negative areas. The first example has all positive y-values while the second has all negative y-values and the third has a mixture of both.(4 votes)

- I don't get why it was F(5 + i/2). I know that it is the height, but how did he get that expression?(3 votes)
- It's actually f(-5+i/2) (emphasis on the negative five). It is set up this way because the rectangles in the graph start at x = -5 and increment by 1/2 for each rectangle. Because it's a right-handed rectangle, the y value that determines the rectangle's height is on the right hand side. So the first rectangle's height is given by the y-value f(-5 + 1/2), the second recangle's by f(-5 +2/2), the third by f(-5 + 3/2), and so on. Notice that the expression says "the value of f at negative 5 (where the rectangles start) plus 1/2 times whichever rectangle we're currently on."

If it's the iteration from i = 1 through i = 8 that is confusing, look into videos on sigma notation. That should help clear it up.(3 votes)

- If this graph were said to describe the velocity of an object against time, could it be said that the absolute value of the area (taking the value of the integral between x=-5 and x=-1 and adding it to the absolute value of the integral between x=-1 and x=7) is equal to the total distance travelled, but the value of area (value of the integral between x=-5 and x=7, so subtraction would occur) is equal to the displacement?(3 votes)
- So why can't area be negative?(2 votes)
- What would be the area of the shaded region in this graph? https://www.desmos.com/calculator/yv4ah6svvu

If we allow negative area, then the answer is 0, does that seem right to you?

It is important to remember that area is just one interpretation of the result of integration. It is commonly used with beginners because we all have experience with the concept of area in the physical sense.

Check out this interpretation of Area Under the Curve:

http://sepia.unil.ch/pharmacology/index.php?id=66

https://en.wikipedia.org/wiki/Area_under_the_curve_(pharmacokinetics)

Keep Studying and

Keep Asking Questions!

.(3 votes)

- Sounds like an unfair question. We might consider negative areas as having a decrease in the antiderivative function.(3 votes)
- At1:35, when we count do we always count starting from the right side if the rectangle?

Because if you see, when you count from the left side of the first blue rectangle, the total count is 9. But if i start counting from the right side of the firzt blue triangle it adds up to be 8.

Can someone please explain this to me(1 vote)- There are 8 rectangles, counting either from the left or from the right.

I guess you're counting the sides rather than the rectangles. By doing so, you'll get 9 ''walls''. Try counting the upper side of each rectangle...(4 votes)

## Video transcript

- [Voiceover] The graph
of F is shown below. A total of 24 right hand rectangles are shown. So, what do I mean by
right hand rectangles? So, there's clearly 24 rectangles. You can count them. And right hand rectangle means that for each of these rectangles the height of the rectangle is defined by the value of the function on the right hand side of the rectangle. So you can see this is the right hand side of this first rectangle and if you take the value of the function of that point that is the height of the rectangle. A left hand rectangle would define the height of the rectangle by the value of the function on the left hand side of the rectangle. So, a left handed rectangle's height, the first rectangle's height would look like that. That's what they mean by right handed rectangle Fair enough, eight in blue. We see that. 16 in red. All right. All 24 of the rectangles
have the same width. Which of the statements below is or are true? They give us three expressions in sigma notation and they say, like this first one is the sum of the areas of the blue rectangles. This is the sum of the areas of the red rectangles. This is the sum of the areas of all of the rectangles. So, I encourage you now to pause the video and try to
determine on your own, which of the statements is or are true. So, I assume you've had a go at it. Let's just go through each of these and see whether they make sense. So this first one, the sum of the areas of the blue rectangles. Well, we know we have one, two, three, four, five, six, seven,
eight blue rectangles, and we're summing from one to eight. So it seems like we're summing eight things right over here. This is one, two, three, four, five, six, seven, eight. So, this is looking good right over here. And then we take F of something times one half. So, we're not even looking at this yet. It looks like this would be the height of each of the rectangles. Remember, we're taking the value of the function on the right hand side for the height, and this would be the width. So does it make sense that the width of each of these rectangles is one half? Well, the total distance between X equals negative five and X equals seven, that distance is 12. Five plus seven, that's 12, and we're dividing it into 24 rectangles of equal width. So, if you divide 12 divided by 24 each of these are going to have a width of one half. So this is checking out that the one half. Now let's think about this part. Let's think about the F of negative five plus I over two. So, let's see. When I is equal to one, so we're going to take one half times F of negative five plus one over two. Right? I is one. So negative five plus one half is going to get us to this point right over here. F of that is going to be this distance, this height right over here. This is consistent with these being right handed rectangles. So, this is definitely the case. When I is equal to one we're definitely finding this
area right over here. When I is equal to two it's going to be negative five plus two over two. So two over two, we're going to add one, we're going to go over here. So once again we're doing one half, which is this right over here. That's the width times F of negative five plus two over two, which is F of negative four, which is this
height right over here. So, once again that is that area. You can keep following it. Everytime we're taking the function this first one is negative five plus one half, and then for each increment we're adding a half to I guess the right hand side is one way to think about it. So, this actually makes complete sense. We're doing this for the first eight so this is indeed true. This is the sum of the areas of the blue rectangles. Now let's look at this one over here. The sum of the areas
of the red rectangles. At first this looks pretty interesting. We're taking the sum of 16 things, and we do indeed have 16 things right over here. We have the width of each of those 16 things or for each of those things we want to figure out an area, and it is indeed the case that each of these has a width of one half. But what happens when we take F of negative one plus I over two. So, we're starting right over here at negative one. Negative one plus I over two. When I equals one we're going to be at this point right over here, and F of that is going to be -- You might say, "Hey, isn't that going to be the height of that rectangle? When I equals two isn't it going to be the height of that rectangle? And when I is equal to three isn't it going to be the height of that rectangle?" And that's where we
have to be very careful. They're going to have the same absolute value, but these are going to be these are all going to be negative values. These are all going to be negative because we see between this value of our function. So that looks like between negative one half all the way to seven our function is actually negative. So one way to think about it, you'd be having negative height so when you multiply these two things you're going to get a negative number. So this whole thing is going to be a negative number and so you're essentially going to get
the negative of the sums of the areas of the red rectangles, but that's not the same thing as the sum of the areas of the red rectangles. An area is, at least in
the traditional sense, you would expect to have, if you were just seeing what's the
area how much carpet you would need to cover
this someone would say that will be a positive value. But this is going to be a negative version of that. So that is not the sum of the areas of the areas of the red rectangles. It's the negative of the areas of the red rectangles. So we'd rule that one out. Then this last choice. So, this expression is the sum of the areas of all the rectangles. And so this one is going from I equals one to 24. So it's literally 24 things. It's starting here and
it just keeps going. And if this said from I
equals one to I equals eight it would be the first choice, but then this falls into the problem again of once we get
past, when we get to I equal nine this thing right over here will turn negative, and it's going to give the negative area. So it's essentially going to net out this positive area against this negative area right over here. So it's not the sum of the areas of all the rectangles. It's going to be this area essentially minus this
area right over there.