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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(x) from x=0 to x=8 using a right Riemann sum with 3 unequal subdivisions.
x0348
f(x)25711
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 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(x) from x=3 to x=11 using a trapezoidal sum with 4 equal subdivisions.
x357911
h(x)364812
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 to try more problems like this? Check out this exercise.

Want to join the conversation?

  • leafers seed style avatar for user Darren Leung
    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)
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    • primosaur ultimate style avatar for user Julia Pockat
      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)
  • duskpin ultimate style avatar for user kathrynjade777
    So do we use Riemann Sums to do numerical integration?
    (3 votes)
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    • leafers seed style avatar for user Travis Bartholome
      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)
  • purple pi teal style avatar for user Julia
    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)
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    • female robot grace style avatar for user loumast17
      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)
  • blobby green style avatar for user ncedwards1234
    Is there ever any difference in area between the trapezoidal and midpoint methods?
    (3 votes)
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    • leaf green style avatar for user kubleeka
      Often, yes. Consider the area under x² from 0 to 1 with 3 subdivisions.

      Midpoint method gives ((1/6)²+(1/2)²+(5/6)²)·(1/3)=35/96
      Trapezoid method gives (1/2)·(1/3)·(0²+(1/3)²+(1/3)²+(2/3)²+(2/3)²+1²)=19/48
      (7 votes)
  • blobby green style avatar for user 100026616
    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)
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    • male robot donald style avatar for user Venkata
      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)
  • blobby green style avatar for user Michael Greer
    For instance ,if n=5, then the area is approximately
    (1 vote)
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  • blobby green style avatar for user djy3forever
    whats the formula for rimann sums that doesn't just use x underscore whatever
    (1 vote)
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