If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Left & right Riemann sums

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.
Problem 1
What kind of Riemann sum is described by the diagram?
Choose 1 answer:

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 subdivisionsNonuniform subdivisions
Problem 2
What is the correct description of the subdivisions in this Riemann sum.
Choose 1 answer:

Riemann sum problems with graphs

Imagine we're asked to approximate the area between y=g(x) and the x-axis from x=2 to x=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 units2, which is an approximation for the area under the curve.
Problem 3
Approximate the area between y=h(x) and the x-axis from x=2 to x=4 using a right Riemann sum with three equal subdivisions.
Choose 1 answer:

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=1 to x=10 using a right Riemann sum with three equal subdivisions. To do that, we are given a table of values for f.
x14710
f(x)6835
A good first step is to figure out the width of each subdivision. The width of the entire area we are approximating is 101=9 units. If we're using three equal subdivisions, then the width of each rectangle is 9÷3=3.
From there, we need to figure out the height of each rectangle. Our first rectangle sits on the interval [1,4]. Since we are using a right Riemann sum, its top-right vertex should be on the curve where x=4, so its y-value is f(4)=8.
In a similar way we can find that the second rectangle, that sits on the interval [4,7], has its top-right vertex at f(7)=3.
Our third (and last) rectangle has its top-right vertex at f(10)=5.
Now all that remains is to crunch the numbers.
First rectangleSecond rectangleThird rectangle
Width333
Height835
Area38=2433=935=15
Then, after finding the individual areas, we'd add them up to get our approximation: 48 units2.
Problem 4
Approximate the area between the x-axis and y=g(x) from x=10 to x=16 using a left Riemann sum with three equal subdivisions.
x10121416
g(x)5177
The approximate area is
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
units2.

Now imagine we're asked to approximate the area between the x-axis and the graph of f(x)=2x from x=3 to x=3 using a right Riemann sum with three equal subdivisions.
The entire interval [3,3] is 6 units wide, so each of the three rectangles should be 6÷3=2 units wide.
The first rectangle sits on [3,1], so its height is f(1)=21=0.5. Similarly, the height of the second rectangle is f(1)=21=2 and the height of the third rectangle is f(3)=23=8.
First rectangleSecond rectangleThird rectangle
Width222
Height0.528
Area20.5=122=428=16
So our approximation is 21 units2.
Problem 5
Approximate the area between the x-axis and h(x)=3x from x=0 to x=1.5 using a right Riemann sum with 3 equal subdivisions.
The approximate area is
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
units2.

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).
Problem 6
Is this Riemann sum an overestimation or underestimation of the actual area?
Choose 1 answer:

Problem 7
Consider the left and right Riemann sums that would approximate the area under y=g(x) between x=2 and x=8.
Are the approximations overestimations or underestimations? Fill in the blanks.
The left Riemann sum would be entirely
the curve, so it would be an
.
The right Riemann sum would be entirely
the curve, so it would be an
.

Problem 8
The continuous function g is graphed.
We're interested in the area under the curve between x=7 and x=7, and we're considering using Riemann sums to approximate it.
Order the areas from least (on top) to greatest (on bottom).
1

Problem 9
This table gives select values of the continuous and increasing function g.
x2381318
g(x)1319283141
We're interested in the area under the curve between x=2 and x=18, and we're considering using left and right Riemann sums, each with four equal subdivisions, to approximate it.
Order the areas from least (on top) to greatest (on bottom).
1

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 sumRight 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.
DirectionLeft Riemann sumRight Riemann sum
IncreasingUnderestimationOverestimation
DecreasingOverestimationUnderestimation

Want to join the conversation?

  • leafers seedling style avatar for user Michael Sands
    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?
    (58 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Tirthankar
      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.
      (80 votes)
  • winston baby style avatar for user Jonathan
    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.
    (10 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Garret Cervantez
      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.
      (21 votes)
  • spunky sam blue style avatar for user Glucogeno
    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)
    Default Khan Academy avatar avatar for user
    • primosaur seed style avatar for user Ian Pulizzotto
      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.
      (9 votes)
  • piceratops ultimate style avatar for user jiantw
    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
    (6 votes)
    Default Khan Academy avatar avatar for user
  • aqualine seed style avatar for user Helen Chen
    why does a right-hand sum underestimate a decreasing graph?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user loumast17
      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)
  • piceratops seed style avatar for user Philippe Nory McCutchan
    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)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user jessieanngough
    Is there an actual formula for finding the right and left Riemann sums?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user BB8FN2187
    Is there a way to report errors in an article? (There is one in this article.)
    (1 vote)
    Default Khan Academy avatar avatar for user
  • female robot grace style avatar for user gcooper7796
    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)
    Default Khan Academy avatar avatar for user
    • blobby blue style avatar for user ongjj
      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)
  • blobby green style avatar for user Susarla, Deepthi
    How to calculate when asked to find by how much the estimated area is overestimated or underestimated?
    (3 votes)
    Default Khan Academy avatar avatar for user