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?

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*.

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.

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.

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

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

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?

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.

why does a right-hand sum underestimate a decreasing graph?

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?

Is there an actual formula for finding the right and left Riemann sums?

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.

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?

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.

How to calculate when asked to find by how much the estimated area is overestimated or underestimated?

actual area minus area calculated using reimann sums. These types of questions are usually not asked.

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?

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

Is there a way to report errors in an article? (There is one in this article.)

Click on the "Ask a question" box, and click "Report a mistake".

Main content

Left & right Riemann sums

Google Classroom

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?

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
The shaded area below the curve is divided into 3 rectangles of equal width. Each rectangle moves upward from the x-axis and touches the curve at the top left corner.	The shaded area below the curve is divided into 3 rectangles of unequal width. Each rectangle moves upward from the x-axis and touches the curve at the top left corner.

Problem 2

What is the correct description of the subdivisions in this Riemann sum.

Riemann sum problems with graphs

Imagine we're asked to approximate the area between

y = g (x)

and the

x

-axis from

x = 2

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

units

^{2}

, which is an approximation for the area under the curve.

For this problem, we can just count the unit squares for each rectangle, but in general we'd need to multiply heights and widths to find area.

	First rectangle	Second rectangle	Third rectangle	Fourth rectangle
Width	$1$ ‍	$1$ ‍	$1$ ‍	$1$ ‍
Height	$3$ ‍	$7$ ‍	$6$ ‍	$4$ ‍
Area	$1 \cdot 3 = 3$ ‍	$1 \cdot 7 = 7$ ‍	$1 \cdot 6 = 6$ ‍	$1 \cdot 4 = 4$ ‍

Then, after finding the individual areas, we'd add them up to get

20

units

^{2}

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.

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

x = 10

using a right Riemann sum with three equal subdivisions. To do that, we are given a table of values for

f


$x$ ‍	$1$ ‍	$4$ ‍	$7$ ‍	$10$ ‍
$f (x)$ ‍	$6$ ‍	$8$ ‍	$3$ ‍	$5$ ‍

A good first step is to figure out the width of each subdivision. The width of the entire area we are approximating is

10 - 1 = 9

units. If we're using three equal subdivisions, then the width of each rectangle is

9 \div 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 rectangle	Second rectangle	Third rectangle
Width	$3$ ‍	$3$ ‍	$3$ ‍
Height	$8$ ‍	$3$ ‍	$5$ ‍
Area	$3 \cdot 8 = 24$ ‍	$3 \cdot 3 = 9$ ‍	$3 \cdot 5 = 15$ ‍

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

48

units

^{2}

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.


$x$ ‍	$10$ ‍	$12$ ‍	$14$ ‍	$16$ ‍
$g (x)$ ‍	$5$ ‍	$1$ ‍	$7$ ‍	$7$ ‍

The approximate area is

units

^{2}

Now imagine we're asked to approximate the area between the

x

-axis and the graph of

f (x) = 2^{x}

from

x = - 3

x = 3

using a right Riemann sum with three equal subdivisions.

The entire interval

[- 3, 3]

6

units wide, so each of the three rectangles should be

6 \div 3 = 2

units wide.

The first rectangle sits on

[- 3, - 1]

, so its height is

f (- 1) = 2^{- 1} = 0.5

. Similarly, the height of the second rectangle is

f (1) = 2^{1} = 2

and the height of the third rectangle is

f (3) = 2^{3} = 8

	First rectangle	Second rectangle	Third rectangle
Width	$2$ ‍	$2$ ‍	$2$ ‍
Height	$0.5$ ‍	$2$ ‍	$8$ ‍
Area	$2 \cdot 0.5 = 1$ ‍	$2 \cdot 2 = 4$ ‍	$2 \cdot 8 = 16$ ‍

So our approximation is

21

units

^{2}

Problem 5

Approximate the area between the $x$ ‍-axis and $h (x) = \frac{3}{x}$ ‍ from $x = 0$ ‍ to $x = 1.5$ ‍ using a right Riemann sum with $3$ ‍ equal subdivisions.

The approximate area is

units

^{2}

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?

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).

Problem 9

This table gives select values of the continuous and increasing function

g

$x$ ‍	$- 2$ ‍	$3$ ‍	$8$ ‍	$13$ ‍	$18$ ‍
$g (x)$ ‍	$13$ ‍	$19$ ‍	$28$ ‍	$31$ ‍	$41$ ‍

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).

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
The graph of the function has the region under the curve divided into 4 rectangles of equal width, touching the curve at the top left corners.	The graph of the function has the region under the curve divided into 4 rectangles of equal width, touching the curve at the top right corners.

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

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Sort by:

Michael Sands
Posted 7 years ago. Direct link to Michael Sands's post “If you take the left and ...”
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?
Button navigates to signup pageComment on Michael Sands's post “If you take the left and ...”
(53 votes)
Answer
- Tirthankar
  Posted 6 years ago. Direct link to Tirthankar's post “If you take the left and ...”
  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.
  Button navigates to signup page
  (68 votes)
Jonathan
Posted 7 years ago. Direct link to Jonathan's post “Ok I still don't understa...”
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.
Button navigates to signup pageButton navigates to signup page
(10 votes)
Answer
- Garret Cervantez
  Posted 6 years ago. Direct link to Garret Cervantez's post “Unequal subdivisions have...”
  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.
  Button navigates to signup page
  (20 votes)
Glucogeno
Posted 6 years ago. Direct link to Glucogeno's post “I don't have clear the qu...”
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?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Ian Pulizzotto
  Posted 6 years ago. Direct link to Ian Pulizzotto's post “The 3 equal subintervals ...”
  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.
  Comment on Ian Pulizzotto's post “The 3 equal subintervals ...”
  (9 votes)
jiantw
Posted 5 years ago. Direct link to jiantw's post “Can a Riemann sum be used...”
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
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- cossine
  Posted 5 years ago. Direct link to cossine's post “Yes it can. Just beware t...”
  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
  Button navigates to signup page
  (4 votes)
Helen Chen
Posted 4 years ago. Direct link to Helen Chen's post “why does a right-hand sum...”
why does a right-hand sum underestimate a decreasing graph?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- loumast17
  Posted 4 years ago. Direct link to loumast17's post “With a right hand sum the...”
  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?
  Button navigates to signup page
  (4 votes)
Philippe Nory McCutchan
Posted 6 years ago. Direct link to Philippe Nory McCutchan's post “If on of the problems was...”
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?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- ggadget6
  Posted 6 years ago. Direct link to ggadget6's post “The problem is similar, b...”
  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
  Button navigates to signup page
  (6 votes)
jessieanngough
Posted 4 years ago. Direct link to jessieanngough's post “Is there an actual formul...”
Is there an actual formula for finding the right and left Riemann sums?
Button navigates to signup pageComment on jessieanngough's post “Is there an actual formul...”
(4 votes)
Answer
- kubleeka
  Posted 4 years ago. Direct link to kubleeka's post “No. Generally, computing ...”
  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.
  Button navigates to signup page
  (2 votes)
BB8FN2187
Posted 5 years ago. Direct link to BB8FN2187's post “Is there a way to report ...”
Is there a way to report errors in an article? (There is one in this article.)
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Alex
  Posted 5 years ago. Direct link to Alex's post “Click on the "Ask a quest...”
  Click on the "Ask a question" box, and click "Report a mistake".
  Button navigates to signup page
  (6 votes)
gcooper7796
Posted 6 years ago. Direct link to gcooper7796's post “Using a Left Riemann sum ...”
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?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- ongjj
  Posted 6 years ago. Direct link to ongjj's post “If the function is below ...”
  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.
  Button navigates to signup page
  (2 votes)
Susarla, Deepthi
Posted 2 years ago. Direct link to Susarla, Deepthi's post “How to calculate when ask...”
How to calculate when asked to find by how much the estimated area is overestimated or underestimated?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Vidit Jain
  Posted 2 years ago. Direct link to Vidit Jain's post “actual area minus area ca...”
  actual area minus area calculated using reimann sums. These types of questions are usually not asked.
  Button navigates to signup page
  (2 votes)

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 6

Left and right Riemann sums

Riemann sum subdivisions/partitions

Riemann sum problems with graphs

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

Riemann sums sometimes overestimate and other times underestimate

Key points to remember

Approximating area under a curve with rectangles

Better approximation with more subdivisions

Left vs. right Riemann sums

Overestimation and underestimation

Want to join the conversation?