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### Course: APยฎ๏ธ/College Calculus BCย >ย Unit 6

Lesson 2: Approximating areas with Riemann sums

# Riemann sums review

Review how we use Riemann sums and the trapezoidal rule to approximate an area under a curve.

## What are Riemann sums?

A Riemann sum is an approximation of the area under a curve by dividing it into multiple simple shapes (like rectangles or trapezoids).
In a left Riemann sum, we approximate the area using rectangles (usually of equal width), where the height of each rectangle is equal to the value of the function at the left endpoint of its base.
In a right Riemann sum, the height of each rectangle is equal to the value of the function at the right endpoint of its base.
In a midpoint Riemann sum, the height of each rectangle is equal to the value of the function at the midpoint of its base.
We can also use trapezoids to approximate the area (this is called trapezoidal rule). In this case, each trapezoid touches the curve at both of its top vertices.
For each type of approximation, the more shapes we use, the closer the approximation would be to the actual area.
Resources differ on this point, but we call any approximation that uses rectangles a Riemann sum, and any approximation that uses trapezoids a trapezoidal sum.
Want to learn more about Riemann sums? Check out this video.

## Practice set 1: Approximating area using Riemann sums

Problem 1.1
Approximate the area between the $x$-axis and $f\left(x\right)$ from $x=0$ to $x=8$ using a right Riemann sum with $3$ unequal subdivisions.
$x$$0$$3$$4$$8$
$f\left(x\right)$$2$$5$$7$$11$
The approximate area is
units${}^{2}$.

Want to try more problems like this? Check out this exercise.

## Practice set 2: Approximating area using the trapezoidal rule

Problem 2.1
Approximate the area between the $x$-axis and $h\left(x\right)$ from $x=3$ to $x=11$ using a trapezoidal sum with $4$ equal subdivisions.
$x$$3$$5$$7$$9$$11$
$h\left(x\right)$$3$$6$$4$$8$$12$
The approximate area is
units${}^{2}$.

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• What is the advantage of Riemann sum over integration when it comes to application?
Cuz I think using integrals to find areas is more convenient and accurate, where I don't need to add up a whole set of numbers to obtain the area.
(27 votes)
• Integration is an easy way to find the area under a graph when you are given a function. However, when you have a set of data points that do not form an obvious equation for a function, integration becomes harder, because you would not have an equation to integrate. Therefore, in a situation where you are given a set of data points that do no the form an obvious equation for a function, it would be easier just to use Riemann sums to determine area.
(127 votes)
• So do we use Riemann Sums to do numerical integration?
(3 votes)
• They're one way to approximate integrals numerically, but typically numerical integration is done with methods that are significantly more accurate than Riemann sums - for example, Simpson's Rule or trapezoidal sums. I believe there are videos about both of those methods here on Khan Academy.
(16 votes)
• I've seen FRQ questions that ask you to justify why these approximations are over/underestimates - what information would we need to include for full points for the justification?
(4 votes)
• with riemann sums it kind of depends on the shape of the graph if it is an over or under estimate, but at the very least talking about if it is increasing or decreasing and if you are doing right sided or left sided sums would normally determine things.

To really get specific though you could talk about if the opposite sides of your rectangles are above or below the graph. so if you are doing right sided rectangles check the left side and see if they are above or below the graph, and if you are doing left sided rectangles check where the upper right corner is in comparison to the graph.

Usually increasing/ decreasing graph and left/right side rectangles is enough though.
(8 votes)
• Is there ever any difference in area between the trapezoidal and midpoint methods?
(3 votes)
• Often, yes. Consider the area under xยฒ from 0 to 1 with 3 subdivisions.

Midpoint method gives ((1/6)ยฒ+(1/2)ยฒ+(5/6)ยฒ)ยท(1/3)=35/96
Trapezoid method gives (1/2)ยท(1/3)ยท(0ยฒ+(1/3)ยฒ+(1/3)ยฒ+(2/3)ยฒ+(2/3)ยฒ+1ยฒ)=19/48
(7 votes)
• So is it that for the right riemann sum you don't include the first term (given a set of data points) and for the left riemann sum you don't include the last? or is it the opposite?
(3 votes)
• You're right. When you take right Riemann sums, you exclude the first data point and when you take left Riemann sums, you exclude the last one.

Frankly speaking, you don't need to memorize this. If you can visualise Riemann sums well enough, then you can come to conclusions without memorized rules.
(5 votes)
• For instance ,if n=5, then the area is approximately
(1 vote)
• if n=5 then we devide the range(which is *X*last-*X*first) by five and that becomes our rectangle base(for riemann sums) or becomes our height(for trapezoidal sums)
(1 vote)
• whats the formula for rimann sums that doesn't just use x underscore whatever
(1 vote)
• sum from k=1 to k=n of (f(a+((k-1)+p)dx)dx) a is the lower limit, b is the upper limit, n is number of rectangles, and p is 0 for left sum, 0.5 for mid sum, and 1 for right sum
(1 vote)