Main content

### Course: Calculus 2 > Unit 1

Lesson 2: Approximation with Riemann sums- Riemann approximation introduction
- Over- and under-estimation of Riemann sums
- Left & right Riemann sums
- Worked example: finding a Riemann sum using a table
- Left & right Riemann sums
- Worked example: over- and under-estimation of Riemann sums
- Over- and under-estimation of Riemann sums
- Midpoint sums
- Trapezoidal sums
- Understanding the trapezoidal rule
- Midpoint & trapezoidal sums
- Riemann sums review
- Motion problem with Riemann sum approximation

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Motion problem with Riemann sum approximation

Here we see how estimating the area under a curve representing velocity estimates distance traveled. Created by Sal Khan.

## Want to join the conversation?

- In short, the more rectangles we come up, the better approximation we can get? But why
**only rectangles**, not**rectangles & triangles**? For example: We all know`Δx = x₁ - x₀`

and the area of the first rectangle is`A = f(x₁) * Δx`

. How about`A = f(x₀) * Δx + 1/2 * [f(x₁) - f(x₀)] * Δx`

?

P/S: If I simplify that hairy expression, it becomes`A = Δx * [1/2 * (f(x₀) + f(x₁))]`

.(46 votes)- Good thinking. Actually, that is one of the formulas. More specifically, the trapezoidal area formula. If you rewrite your formula, it is height*(1/2(base 1 + base 2)).(46 votes)

- If the graph was showing the acceleration instead of the velocity, would the area under the curve still be the distance traveled?(5 votes)
- No, the area under the curve would be the velocity.(23 votes)

- Would the average of right rectangles and left rectangles be a more accurate approximation than the two individually?(3 votes)
- if you take right handed and left handed rectangles then take average of those two, would you get an exact value of the total distance traveled?(2 votes)
- I would recommend looking at a graph while doing this. On your graph, pick a x sub 0 and a x sub 1 (as your width.) Then draw a rectangle with it's height being the average height of that graph within your interval. When we look for the exact area under a curve, we don't want our rectangle to be above or below the curve at any point. As you can see when you draw the rectangle, parts of it exceeds the curve and parts of it under exceeds the curve. Taking the average of the heights for your rectangle would give you a better approximation of the actually area under the curve, but it would still not give you the actually area under the curve.(4 votes)

- Wouldn't it be more accurate to do this with triangles and trapezoids? Why don't we do it like that?(3 votes)
- Yes it is - and you will get there in a few more videos.

Sneak Peek: https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/riemann-sums/v/rectangular-and-trapezoidal-riemann-approximations(1 vote)

- Would you get the exact distance traveled if you took the average of the over estimation plus the under estimation?(1 vote)
- Just image a parabula y = x^2 from 0 to 1:

Overestimate with 1*1= 1 and underestimae with 0*1=0

So the average would be 1/2(0+1) = 1/2 , wich is the exact area of a triangle and not the curve under the parabula.(3 votes)

- If the right riemann sum is an overestimate of the area why don't we take the limit of the curve to precisely approximate the area under the curve like we did in derivatives. Please correct me if I am wrong.(1 vote)
- In this example we can't take the limit of the curve, because we don't have a continuous function for the curve: we only know what the velocity is at two second intervals, not what happens in between. However, if you do have a continuous function, then yes, you can decrease the width of the rectangles to the limit of width = 0. That's where this tutorial is heading.(3 votes)

- Is there a way to caculate the exact number of the area?(2 votes)
- Yes. That's what the rest of the integral calculus playlist is about. The gist is that we take the limit of a sequence of better and better approximations.(2 votes)

- In this specific problem, could you take both a left and a right-handed limit, find the delta, subtract it from the left-handed limit, and get something more exact?(1 vote)
- Yes, you can use the left and right-hand values (not limits) and use them to calculate a somewhat more accurate approximation. This is called the trapezoid method, and Sal covers it later in this sequence.(4 votes)

- If right-handed rectangles over-estimate the area, and left-handed rectangles under-estimate the area, the to get the most accurate approximation, can we use the mid-point?(1 vote)
- Yes, midpoint is generally a more accurate approximation than left-hand or right-hand rectangle approximations. In the midpoint approximation, the height of each rectangle is the value of the function at the midpoint of the interval.

Have a blessed, wonderful day!(3 votes)

## Video transcript

Voiceover: A cyclist starts pedaling and accelerates for 12 seconds. The velocity v of t of the cyclist at two second intervals in feet per second is given by the table. So they tell us at different times. After four seconds the velocity is 7.5 feet per second. After eight seconds the velocity is nine feet per second. Consider the graph of
velocity versus time. Velocity versus time. Let capital r of six be
the sum of the areas of six right hand rectangles
with equal sub-divisions. It follows that capital r of six is an approximation for the total distance traveled in feet during the 12 seconds. What is the value of capital r of six? I encourage you to pause this video and try to think about it on your own and then I'll work through it and at any time while I'm working through it, if you get inspired, feel free to pause again and try to take it all the way. Let's think about what they're saying. They say "consider the graph of velocity versus time." We might as well try to plot that. I've got some graph paper here and we can focus on the first quadrant because all of our time values and all of our velocity values are positive. Let's see this is time, this is velocity as a function of time axis, time goes between zero and 12 and they're giving it to us in every two seconds. So, this is zero, two, four, six, eight, ten and 12 seconds, and then our velocity goes between zero and ten feet per second. So, this is one, two, three, four, five, I'll just mark off some of them. Six, seven, eight, nine, ten. Now, this is in feet per second and this over here is in seconds. Let's plot these points at time zero, velocity is zero. Time two velocity is six feet per second. Time four at 7.5 gets us right there. Time six it's 8.5, that 8, 8.5. Time eight at eight seconds it's nine feet per second. At ten seconds it's 9.5 feet per second, and at 12 seconds it's ten feet, it's ten feet per second. What I've done here is I've just plotted the data that they've
given us, considering the graph of velocity
versus time, at least the data they've given us. We can imagine fitting a curve that looks something like this. If we were to, these
points are just sampled from a curve, they might
look something like, something like this, over there. We are now considering the graph of velocity versus time. Now let's think about the sum of the areas of six right handed rectangles with equal sub-divisions. When they're talking
about equal sub-divisions, they're talking about equal sub-divisions along the independent
axis or the time axis in this case and we're
talking about the first 12 seconds, so if we
were to divide the first 12 seconds into six equal
sections, they would each be 2 seconds wide. This would be one of them. Let me do this in a new color. They would each be 2 seconds. That's not a new color, oh it is, blue. They're each going to be two seconds. Let me do one that
contrasts with green better. Each of our rectangles are going to be two seconds wide and I'm just kind of starting them off at the bottom because we have to think about how tall to make them. Then they say to do
right handed rectangles. What's a right handed rectangle? That means we define the height of the rectangle by the value of the function on the right hand side. So, for example this first rectangle, to make it a right handed
rectangle, we look at the right side, we're
at two seconds, velocity at two seconds is six feet per second and so that's going to be the height of our rectangle. This next one is going to look like this. Now you might say "Well, what's a left handed rectangle then?" Well, a left handed
rectangle would be doing something like, would be
doing something like... Okay, for this first rectangle on the left hand side of it, my function is zero. So, it's just zero. Now the next rectangle,
the left hand side, my function at two seconds, the function is six seconds. So, the left hand rectangles would look like this, would look something like that. But anyway they're telling us to do right handed rectangles so let's do right handed rectangles. Let me clear this a little bit. This is our first right handed rectangle. This is our second one. Let me just draw the tops. Third, fourth, fifth and sixth that was the tops of them. It's just going to look
something like this. It's going to look like this. This is my third one. That's my fourth one. That's my fifth one. Then, that's my sixth one. They're tell us let capital r of six be the sum of the areas of these things and they tell us that it follows that capital r of six is an approximation for the total distance traveled in feet during the 12 seconds. Now why is that? Why is the sum of these
areas an approximation for the total distance traveled? Well before calculus, you learned, that if you have a constant rate, that distance is equal to rate times time. Once again, this is assuming, that assumes a constant rate. Our rate in this example
is clearly changing. We are clearly accelerating
right over here but maybe we can approximate the distance traveled if we assume a constant rate over an interval, say over
two second intervals. Then we just, we try to roughly, when we assume some velocity
over that interval. When we take these
right handed rectangles, let's just focus on this first rectangle. If I'm taking the area of this first rectangle, what am I doing? I'm multiplying it's
height times it's width. It's height is going to be the velocity at the end of two seconds and
it's width is the two seconds. If I were to multiply six feet per second times two seconds. That will get me 12 feet,
let me make that very clear. The area for that one area is equal to six feet per second, my
velocity, it's actually the speed but we're not going to get particular about whether we're getting the direction as well. Six feet per second times two seconds. That's going to get me 12 feet. That we can say is an
approximation because we're saying "Okay, you
trialed at two seconds "at some velocity" but is that going to be an overestimate or an underestimate? It's going to be an overestimate because you're taking the fastest velocity over the interval. The interval started with the cyclist going zero feet per second. It's clearly overestimating
and if you think about it and I encourage you to pause the video and think about it on your own. If we did a left handed rectangle it would be underestimating it because we would be taking the slowest
velocity over the interval and multiplying it by the interval. This is an approximation. Let's just calculate what r of six is. R of six is going to be the area of this rectangle which, as we just said, it's six feet per second times two seconds so it's going to be 12
feet and then we have this rectangle which
is going to be 7.5 feet per second times two
seconds which is going to be 15 feet, 15 feet
and you can even see it in the area right over here. It's going to be three feet more. One, two and then half of two of these to it's going to be three feet more and then the area of this one which is going to be 8.5, so our velocity at the end of the interval is going to be 8.5 feet per seconds times two seconds which is 17 feet and then plus this area
which is our velocity at the end of the interval which is nine feet per second times two seconds so plus 18 and then we get the
velocity of this area which is going to be this height. The velocity at the end of the interval which is 9.5 feet per second times two seconds so, plus 19. Then finally this area
which is the velocity at the end of the
interval which is ten feet per second times two seconds so, plus 20. What does this give us? This is the point of which I'm most likely going to make an error. 12 plus 15 is 27 and then 27 plus 10 is 37 plus another seven is
44, so that gives us 44. 44 plus 18 is going to give me, let's see, I can get to 52 and then
62, so this is going to be 62 plus 19 is 81, plus 20 is 101, that's going to be in feet. Capital r of six, 101 feet, it is an approximation for the total distance traveled during the 12 seconds but it's really an over approximation because for every interval, we're taking the fastest velocity of the interval. If we took the slowest
velocity of the interval in would be an underestimate. If we did left handed
rectangles and we might have gotten something in between if we say maybe took the average of these too. If we made the height of the rectangle three for this first one
or something like that. We'll do that in the
future, in future videos.