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# Riemann sums review

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)
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.

## Practice set 1: Approximating area using Riemann sums

Problem 1.1
Approximate the area between the x-axis and f, left parenthesis, x, right parenthesis from x, equals, 0 to x, equals, 8 using a right Riemann sum with 3 unequal subdivisions.
x0348
f, left parenthesis, x, right parenthesis25711
The approximate area is
unitssquared.

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 parenthesis, x, right parenthesis from x, equals, 3 to x, equals, 11 using a trapezoidal sum with 4 equal subdivisions.
x357911
h, left parenthesis, x, right parenthesis364812
The approximate area is
unitssquared.

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.
• 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.
• 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?
• For trapezoidal sums, you would need to specify whether the function was concave up or concave down. If the function (or the part of it that you're approximating) is concave up, like the function f(x)=x^2, then a trapezoidal sum gives an overestimate. If the function is concave down, like g(x)=ln(x), then a trapezoidal sum gives an underestimate.
You don't need that information for rectangular Riemann sums. For them, stick with describing whether the function is increasing or decreasing, as in loumast17's response.
• So do we use Riemann Sums to do numerical integration?
• The trapeziodal sum is exactly that. To visualise this look at the left Riemann sum above and imagine the rectangle above it up to the right Riemann sum. Split this smaller rectangle in half diagonally from bottom left to top right and this gives you the average of the left and right Riemann sums (the area is halfway between both), and it is exactly the trapeziodal form.