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## 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

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# Left & right Riemann sums

AP.CALC:

LIM‑5 (EU)

, LIM‑5.A (LO)

, LIM‑5.A.1 (EK)

, LIM‑5.A.2 (EK)

, LIM‑5.A.3 (EK)

, LIM‑5.A.4 (EK)

Areas under curves can be estimated with rectangles. Such estimations are called Riemann sums.

Suppose we want to find the area under this curve:

We may struggle to find the exact area, but we can approximate it using rectangles:

And our approximation gets better if we use more rectangles:

These sorts of approximations are called

**Riemann sums**, and they're a foundational tool for integral calculus. Our goal, for now, is to focus on understanding two types of Riemann sums: left Riemann sums, and right Riemann sums.## Left and right Riemann sums

To make a Riemann sum, we must choose how we're going to make our rectangles. One possible choice is to make our rectangles touch the curve with their top-left corners. This is called a

**left Riemann sum**.Another choice is to make our rectangles touch the curve with their top-right corners. This is a

**right Riemann sum**.Neither choice is strictly better than the other.

### Riemann sum subdivisions/partitions

Terms commonly mentioned when working with Riemann sums are "subdivisions" or "partitions." These refer to the number of parts we divided the x-interval into, in order to have the rectangles. Simply put, the number of subdivisions (or partitions) is the number of rectangles we use.

Subdivisions can be

**uniform**, which means they are of equal length, or**nonuniform**.Uniform subdivisions | Nonuniform subdivisions |
---|---|

## Riemann sum problems with graphs

Imagine we're asked to approximate the area between y, equals, g, left parenthesis, x, right parenthesis and the x-axis from x, equals, 2 to x, equals, 6.

And say we decide to use a left Riemann sum with four uniform subdivisions.

**Notice:**Each rectangle touches the curve at its top-left corner because we're using a

*left*Riemann sum.

Adding up the areas of the rectangles, we get 20 unitssquared, which is an approximation for the area under the curve.

## Now let's do some approximations without the aid of graphs.

Imagine we're asked to approximate the area between the x-axis and the graph of f from x, equals, 1 to x, equals, 10 using a right Riemann sum with three equal subdivisions. To do that, we are given a table of values for f.

x | 1 | 4 | 7 | 10 | |

f, left parenthesis, x, right parenthesis | 6 | 8 | 3 | 5 |

A good first step is to figure out the width of each subdivision. The width of the entire area we are approximating is 10, minus, 1, equals, 9 units. If we're using three equal subdivisions, then the width of each rectangle is 9, divided by, 3, equals, start color #11accd, 3, end color #11accd.

From there, we need to figure out the height of each rectangle. Our first rectangle sits on the interval open bracket, 1, comma, 4, close bracket. Since we are using a

**right**Riemann sum, its top-right vertex should be on the curve where x, equals, 4, so its y-value is f, left parenthesis, 4, right parenthesis, equals, start color #e07d10, 8, end color #e07d10.In a similar way we can find that the second rectangle, that sits on the interval open bracket, 4, comma, 7, close bracket, has its top-right vertex at f, left parenthesis, 7, right parenthesis, equals, start color #7854ab, 3, end color #7854ab.

Our third (and last) rectangle has its top-right vertex at f, left parenthesis, 10, right parenthesis, equals, start color #ca337c, 5, end color #ca337c.

Now all that remains is to crunch the numbers.

First rectangle | Second rectangle | Third rectangle | |
---|---|---|---|

Width | start color #11accd, 3, end color #11accd | start color #11accd, 3, end color #11accd | start color #11accd, 3, end color #11accd |

Height | start color #e07d10, 8, end color #e07d10 | start color #7854ab, 3, end color #7854ab | start color #ca337c, 5, end color #ca337c |

Area | start color #11accd, 3, end color #11accd, dot, start color #e07d10, 8, end color #e07d10, equals, 24 | start color #11accd, 3, end color #11accd, dot, start color #7854ab, 3, end color #7854ab, equals, 9 | start color #11accd, 3, end color #11accd, dot, start color #ca337c, 5, end color #ca337c, equals, 15 |

Then, after finding the individual areas, we'd add them up to get our approximation: 48 unitssquared.

Now imagine we're asked to approximate the area between the x-axis and the graph of f, left parenthesis, x, right parenthesis, equals, 2, start superscript, x, end superscript from x, equals, minus, 3 to x, equals, 3 using a right Riemann sum with three equal subdivisions.

The entire interval open bracket, minus, 3, comma, 3, close bracket is 6 units wide, so each of the three rectangles should be 6, divided by, 3, equals, start color #11accd, 2, end color #11accd units wide.

The first rectangle sits on open bracket, minus, 3, comma, minus, 1, close bracket, so its height is f, left parenthesis, minus, 1, right parenthesis, equals, 2, start superscript, minus, 1, end superscript, equals, start color #e07d10, 0, point, 5, end color #e07d10. Similarly, the height of the second rectangle is f, left parenthesis, 1, right parenthesis, equals, 2, start superscript, 1, end superscript, equals, start color #7854ab, 2, end color #7854ab and the height of the third rectangle is f, left parenthesis, 3, right parenthesis, equals, 2, cubed, equals, start color #ca337c, 8, end color #ca337c.

First rectangle | Second rectangle | Third rectangle | |
---|---|---|---|

Width | start color #11accd, 2, end color #11accd | start color #11accd, 2, end color #11accd | start color #11accd, 2, end color #11accd |

Height | start color #e07d10, 0, point, 5, end color #e07d10 | start color #7854ab, 2, end color #7854ab | start color #ca337c, 8, end color #ca337c |

Area | start color #11accd, 2, end color #11accd, dot, start color #e07d10, 0, point, 5, end color #e07d10, equals, 1 | start color #11accd, 2, end color #11accd, dot, start color #7854ab, 2, end color #7854ab, equals, 4 | start color #11accd, 2, end color #11accd, dot, start color #ca337c, 8, end color #ca337c, equals, 16 |

So our approximation is 21 unitssquared.

*Want more practice? Try this exercise.*

## Riemann sums sometimes overestimate and other times underestimate

Riemann sums are approximations of the area under a curve, so they will almost always be slightly more than the actual area (an overestimation) or slightly less than the actual area (an underestimation).

*Want more practice? Try this exercise.*

**Notice:**Whether a Riemann sum is an overestimation or an underestimation depends on whether the function is increasing or decreasing on the interval, and on whether it's a left or a right Riemann sum.

## Key points to remember

### Approximating area under a curve with rectangles

The first thing you should think of when you hear the words "Riemann sum" is that you're using rectangles to estimate the area under a curve. In your mind, you should envision something like this:

### Better approximation with more subdivisions

In general, the more subdivisions (i.e. rectangles) we use to approximate an area, the better the approximation.

### Left vs. right Riemann sums

Try not to mix them up. A left Riemann sum uses rectangles whose top-

*left*vertices are on the curve. A right Riemann sum uses rectangles whose top-*right*vertices are on the curve.Left Riemann sum | Right Riemann sum |
---|---|

### Overestimation and underestimation

When using Riemann sums, sometimes we get an overestimation and other times we get an underestimation. It's good to be able to reason about whether a particular Riemann sum is overestimating or underestimating.

In general, if the function is always increasing or always decreasing on an interval, we can tell whether the Riemann sum approximation will be an overestimation or underestimation based on whether it's a left or a right Riemann sum.

Direction | Left Riemann sum | Right Riemann sum |
---|---|---|

Increasing | Underestimation | Overestimation |

Decreasing | Overestimation | Underestimation |

## Want to join the conversation?

- I don't have clear the question on function 3/x. I get 9 as the answer, but it is incorrect. Maybe I am taking the Riemann sum in the wrong side. Can anyone help me?(5 votes)
- The 3 equal subintervals are [0, 0.5], [0.5, 1], and [1, 1.5], with right-hand endpoints of 0.5, 1, and 1.5. In the right-hand Riemann sum for the function 3/x, the rectangles have heights 3/0.5, 3/1, and 3/1.5; the width of each rectangle is 0.5.

The sum of the areas of these rectangles is 0.5(3/0.5 + 3/1 + 3/1.5) = 5.5, the correct answer.

You might have gotten the answer 9 from missing the last term and failing to multiply by the width of each rectangle, i.e. calculating 3/0.5 + 3/1. However, without seeing your work, I cannot determine for certain what error(s) you made.(8 votes)

- Ok I still don't understand the concept of unequal and equal subdivisions. How are they different? In the practice, I can't seem to get it.(8 votes)
- Unequal subdivisions have a varying distance between the x values. Equal subdivisions have a fixed length between subdivisions. Because the area of a rectangle is the height (y value) times the width (x value) you need to take that into consideration. If you continue to have problems I think that drawing the points on a paper could help you.(2 votes)

- Can a Riemann sum be used to find the exact value of the area under a curve? Ex. if we used infinitely small rectangles to get really close(5 votes)
- Yes it can. Just beware the area of each rectangle is f(x)dx, so if f(x)<0 there area of the rectangle will turn out negative as f(x)dx<0. Note that dx cannot be negative as it is width of each rectangle where dx = lim_x->0 x(3 votes)

- why does a right-hand sum underestimate a decreasing graph?(4 votes)
- With a right hand sum the rectangles meet the line of the graph at their upper right hand corner. Since it is decreasing that means moving to the left the line will move upward on average. This also means, since the rectangle will be under the graph to its left on average, there will be space abov the rectangles that is not measured.

Does that make sense?(4 votes)

- If on of the problems was asking to find the area using equal width trapezoid instead of rectangles, how would be able to solve the problem?(2 votes)
- The problem is similar, but not exactly the same. Your width would be the same. However, instead of multiplying by the leftmost point or the rightmost point in the interval, multiply by the average of the two points. That will give you the area of the trapezoid. If my explanation is confusing or unhelpful, check out this video: https://www.khanacademy.org/math/ap-calculus-ab/ab-accumulation-riemann-sums/ab-midpoint-trapezoid/v/trapezoidal-approximation-of-area-under-curve(6 votes)

- Is there a way to report errors in an article? (There is one in this article.)(2 votes)
- Click on the "Ask a question" box, and click "Report a mistake".(4 votes)

- Using a Left Riemann sum on a increasing function the Riemann method gives us an underestimation...but what if the function is below the X-Axis? The function can still be increasing and negative. Using a Left Riemann sum is the function an still underestimation or overestimation?(3 votes)
- If the function is below the x-axis, then the Riemann sum will be negative. The total area between the endpoints of the interval for some curve is really a net area, where the total area below the x-axis (and above the curve) is subtracted from the

total area above the x-axis (and below the curve).

In the case of the left Riemann sum, the rectangles will "stick out" below the function and be more negative (less) than the actual integral.(2 votes)

- How to calculate when asked to find by how much the estimated area is overestimated or underestimated?(3 votes)
- actual area minus area calculated using reimann sums. These types of questions are usually not asked.(2 votes)

- If you take the left and right Riemann sum then take the average of the two, how close to the actual area will the result be? Also because of the curve, would the result of increasing function always be more than the actual area and opposite for the decreasing?(2 votes)
- If you take the left and right Riemann Sum and then average the two, you'll end up with a new sum, which is identical to the one gotten by the Trapezoidal Rule. (In fact, according to the Trapezoidal Rule, you take the left and right Riemann Sum and average the two.) This sum is more accurate than either of the two Sums mentioned in the article. However, with that in mind, the Midpoint Riemann Sum is usually far more accurate than the Trapezoidal Rule.

To answer the second question, it definitely depends on the concavity of the curve (i..e. whether it's facing up or down). The trapezoidal sum will give you overestimates if the graph is concave up (like y=x^2 + 1) and underestimates if the graph is concave down (like y=-x^2 - 1). Moreover, the Midpoint rule is more accurate than the Trapezoidal rule**given that the concavity does not change**.(2 votes)

- Is there an actual formula for finding the right and left Riemann sums?(2 votes)
- No. Generally, computing infinite sums has to be done on a case-by-case basis, and there are many that don't have a closed form expression.(2 votes)