Riemann Sum: Meaning, Formula, Limit & Method

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Forming Riemann sums area parabola rectangles StudySmarter Approximation of the area below a curve using rectangles - StudySmarter Originals

The first step is to divide the interval of integration into as many subintervals as we want. Remember that the more subintervals we use, the better approximation we get. Right-endpoint and left-endpoint approximations use the very points we obtain when making a partition to find the height of each rectangle. However, we are not limited to these points!

Sigma notation formulas

Before proceeding, we will introduce some formulas in Sigma Notation that will make our lives easier when working with sums.

The following expressions give the sum of consecutive integers, square of integers, and cube of integers, respectively:

$\sum_{i = 1}^{n} i = \frac{n (n + 1)}{2}$

$\sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

$\sum_{i = 1}^{n} i^{3} = {(\frac{n (n + 1)}{2})}^{2}$

The above formulas, along with basic properties of summations, are very useful when approximating areas using sigma notation. Let's see an example using them!

Evaluate the following sum: $S = \sum_{i = 1}^{10} (i^{2} + 2 i - 1)$

Let's use the properties of summation and the above formulas to evaluate the given sum!

Use the properties of summation to rewrite the sum.

$S = \sum_{i = 1}^{10} i^{2} - 2 {\sum i}_{i = 1}^{10} - {\sum 1}_{i = 1}^{10}$

Use $\sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ with $n = 10$ to evaluate the first sum.

$S = \frac{10 (10 + 1) (2 (10) + 1)}{6} - 2 \sum_{i = 1}^{10} i - {\sum 1}_{i = 1}^{10}$

Simplify.

$S = 385 - 2 \sum_{i = 1}^{10} i - \sum_{i = 1}^{10} 1$

Use $\sum_{i = 1}^{10} i = \frac{n (n + 1)}{2}$ with $n = 10$ to evaluate the second sum.

$S = 385 - \frac{10 (10 + 1)}{2} - \sum_{i = 1}^{10} 1$

Simplify.

$S = 385 - 55 - \sum_{i = 1}^{10} 1$

The last sum is just adding 1 together 10 times, which is the same as 10 times 1.

$S = 385 - 55 - 10$

Simplify.

$S = 320$

Forming Riemann Sums

One way to approximate the area below a curve is to divide the area into rectangles. We get this done in the following way:

Begin by dividing the interval of the area into subintervals. The more, the better!
Associate a rectangle to each subinterval.
The width of each rectangle is equal to the length of its corresponding subinterval.
The height of each rectangle can be found by evaluating the function at some point within the subinterval.
- The left-endpoint approximation uses the leftmost value of the subinterval.
- The right-endpoint approximation uses the rightmost value of the subinterval.

But what limits our choice to just the leftmost value or the rightmost value? Nothing! As long as we take any point within the interval, we are good to go!

Of course our approximations will differ, but we should focus on increasing the number of intervals rather than worrying about which point to use for the height of the rectangles.

Riemann sum definition

After dividing the area below a curve into rectangles, we add them up, and we get an approximation for the area. This is known as a Riemann Sum, named after the mathematician Bernhard Riemann, who worked on the idea in the 19th century.

We need to set up some background before defining a Riemann Sum. Let $f (x)$ be a function defined on a closed interval $a \leq x \leq b$ and let $P = {x_{i}}$ for $i = 0, 1, 2, . . ., n$ be a regular partition of $a \leq x \leq b$ . We have a set of n subintervals of the form $x_{i - 1} \leq x \leq x_{i}$ , all with a length of $Δ x = \frac{b - a}{n}$ .

Forming Riemann sums regualar partition StudySmarter A regular partition of the interval - StudySmarter Originals

Up to this point, we have the same set-up for the right-endpoint and left-endpoint approximations using equally-spaced subintervals. The difference comes from which point we use to find the height of the rectangles.

A Riemann Sum for the function $f (x)$ for the partition P is defined as:

$\sum_{i = 1}^{n} f (x_{i}^{*}) Δ x$

Where $x_{i}^{*}$ is any value inside the interval $x_{i - 1} \leq x \leq x_{i}$ , with $x_i \in P$.

The right-endpoint and left-endpoint approximations are particular cases of Riemann Sums.

Riemann sums and midpoint approximation

As we can use any value inside each subinterval when finding Riemann sums, why not use the midpoint? This is known as the midpoint approximation.

The midpoint approximation $M_{n}$ of the area below a curve is a particular case of a Riemann sum obtained by choosing the midpoint of each subinterval. That is:

$M_{n} = \sum_{i = 1}^{n} f (\frac{x_{i - 1} + x_{i}}{2}) Δ x$

As usual, this is better understood with an example. Let's take a look at a midpoint approximation!

Use a midpoint approximation to approximate the area below $f (x) = x^{2} + 1$ in the interval $0 \leq x \leq 2$ by dividing the interval into $10$ subintervals of the same size.

Find the length of each subinterval.

$Δ x = \frac{b - a}{n} = \frac{2 - 0}{10} = 0.2$

Since the length of each subinterval is 0.2, our partition will be made of the points 0, 0.2, 0.4, ..., 1.6, 1.8, and 2. This can be summarized as follows:

$x_{i} = 0.2 i$

We are using the midpoint approximation, so the midpoint between each two values of the partition will be used to find the height of the rectangles.

Use the formula for the midpoint approximation with n=10 and $Δ x = 0.2$ .

$M_{10} = \sum_{i = 1}^{10} f (\frac{x_{i - 1} + x_{i}}{2}) (0.2)$

Rather than writing $\frac{x_{i - 1} + x_{i}}{2}$ we can use the expression for $x_{i}$ and simplify.

$\frac{x_{i - 1} + x_{i}}{2} = \frac{0.2 (i - 1) + 0.2 i}{2} = 0.2 i - 0.1$

We can substitute back this expression into our formula and use the properties of summation to find our approximation.

$M_{10} = \sum_{i = 1}^{10} f (0.2 i - 0.1) (0.2)$

Factor out 0.2 from the summation.

$M_{10} = 0.2 \sum_{i = 1}^{10} f (0.2 i - 0.1)$

Evaluate the function at $0.2 i - 0.1$

$M_{10} = 0.2 \sum_{i = 1}^{10} ({(0.2 i - 0.1)}^{2} + 1)$

Expand the binomial and simplify.

$M_{10} = 0.2 \sum_{i = 1}^{10} (0.04 i^{2} - 0.04 i + 1.01)$

Use the properties of summation to rewrite the sum.

$M_{10} = 0.008 \sum_{i = 1}^{10} i^{2} - 0.008 \sum_{i = 1}^{10} i + 0.202 \sum_{i = 1}^{10} 1$

Use $\sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ to evaluate the first sum.

$M_{10} = 0.008 \frac{10 (11) (21)}{6} - 0.008 \sum_{i = 1}^{10} i + 0.202 \sum_{i = 1}^{10} 1$

Use $\sum_{i = 1}^{n} i = \frac{n (n + 1)}{2}$ to evaluate the second sum.

$M_{10} = 0.008 \frac{10 (11) (21)}{6} - 0.008 \frac{10 (11)}{2} + 0.202 \sum_{i = 1}^{10} 1$

Evaluate the last sum.

$M_{10} = 0.008 \frac{10 (11) (21)}{6} - 0.008 \frac{10 (11)}{2} + 2.02$

Evaluate using a calculator.

$M_{10} = 4.66$

Remember that we get better approximations as we divide the interval into more subintervals!

Until now, we have been approximating areas without knowing which approximation is better. Is there a way of telling which Riemann sum is better? Sadly, if we do not know the real value of the area below the curve, there is no telling which approximation is better. Besides, it would be pointless to do an approximation while knowing the actual value!

However, there is a way of squeezing the value of the area between two values.

Upper Riemann sum

We are allowed to take any value inside each subinterval to form our Riemann sum. What if we take the maximum value of $f$

in each subinterval? The Riemann sum obtained through this process is called an upper Riemann sum.

A Riemann sum for the function $f (x)$ for the partition $P = {x_{i}}$ is called an upper Riemann sum if the values $x_{i}^{*}$ are taken as the maximum value from each subinterval.

In this case, we can guarantee that our approximation will be greater than or equal to the actual value of the area since we are taking bigger pieces of the area! If we let A be the area below the curve and $S_{U}$ be an upper Riemann sum, then we can write the following inequality:

$S_{U} \geq A$

Let's now find an upper Riemann sum of the function of our previous example, $f (x) = x^{2} + 1$ , in the same interval $0 \leq x \leq 2$ by dividing the interval into 10 subintervals as well. We will begin by taking a look at its graph.

Forming Riemann sums parabola StudySmarter Graph of the parabola f(x) divided with a left-endpoint approximation - StudySmarter Originals

We can notice how this function is an increasing function, hence, the greatest value from each subinterval is the rightmost value. Therefore, doing a right-endpoint approximation will give us an upper Riemann sum.

Use the formula for the right-endpoint approximation.

$R_{10} = \sum_{i = 1}^{10} f (x_{i}) Δ x$

Substitute $x_{i} = i Δ x$ and $Δ x = \frac{2 - 0}{10} = 0.2$ into the formula.

$R_{10} = \sum_{i = 1}^{10} f (0.2 i) 0.2$

Factor out 0.2 and evaluate the function.

$R_{10} = 0.2 \sum_{i = 1}^{10} ({(0.2 i)}^{2} + 1)$

Use the properties of summation to rewrite the sum.

$R_{10} = 0.008 \sum_{i = 1}^{10} i^{2} + 0.2 \sum_{i = 1}^{10} 1$

Use $\sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ to evaluate the first sum.

$R_{10} = 0.008 \frac{10 (11) (21)}{6} + 0.2 \sum_{i = 1}^{10} 1$

Evaluate the last sum.

$R_{10} = 0.008 \frac{10 (11) (21)}{6} + 0.2 (10)$

Evaluate using a calculator.

$R_{10} = 5.08$

We can see that the value of our approximation is greater than the midpoint approximation. We also know that this value is greater than the actual area below the curve!

Lower Riemann sum

And what if we take the minimum value of $f$ in

each subinterval? We get a lower Riemann sum!

A Riemann sum for the function $f (x)$ for the partition $P = {x_{i}}$ is called a lower Riemann sum if the values $x_{i}^{*}$ are taken as the minimum value from each subinterval.

This time, we can guarantee that our approximation will be less than or equal to the actual value of the area since we are taking smaller pieces of the area! If we let A be the area below the curve and $S_{L}$ be a lower Riemann sum, then we can write the following inequality:

$S_{L} \leq A$

It's time to find a lower Riemann sum of our previous examples. In this case, the least value from each subinterval is the leftmost value, so a left-endpoint approximation will give us a lower Riemann sum.

Use the formula for the left-endpoint approximation.

$L_{10} = \sum_{i = 0}^{9} f (x_{i}) Δ x$

Substitute $x_{i} = i Δ x$ and $Δ x = \frac{2 - 0}{10} = 0.2$ into the formula.

$L_{10} = \sum_{i = 0}^{9} f (0.2 i) 0.2$

Factor out 0.2 and evaluate the function.

$L_{10} = 0.2 \sum_{i = 0}^{9} ({(0.2 i)}^{2} + 1)$

Use the properties of summation to rewrite the sum.

$L_{10} = 0.008 \sum_{i = 0}^{9} i^{2} + 0.2 \sum_{i = 0}^{9} 1$

Note that the summations start at 0. The last sum still consists of 10 terms, so it can be evaluated as adding 1 ten times together.

Evaluate the last sum.

$L_{10} = 0.008 \sum_{i = 0}^{9} i^{2} + 0.2 (10)$

For the first sum note that the term containing $i = 0$ does not contribute to the sum because $0^{2} = 0$ , so this sum can be evaluated as if it was from $i = 1$ to 9.

Use $\sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ to evaluate the sum.

$L_{10} = 0.008 \frac{9 (10) (19)}{6} + 0.2 (10)$

Evaluate using a calculator.

$L_{10} = 4.28$

We got an approximation of 4.28. This means that the actual value of the area below the curve lies between 4.28 and 5.08!

Riemann sums and area below a curve

We have seen how lower and upper Riemann sums give us a bound on the area below a curve. This also depends on how many subintervals we divide the area below the curve. What if we use an infinite amount of subintervals?

Let $f (x)$ be a continuous function on an interval $a \leq x \leq b$ . Then, the area below the curve is obtained as the limit as n goes to infinity of a Riemann sum:

$A = \lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}^{*}) Δ x$

Where $x_{i}^{*}$ is any value inside the interval $x_{i - 1} \leq x \leq x_{i}$ and $Δ x = \frac{b - a}{n}$ .

Let's think of the above expression: as we increase n, we get thinner rectangles, which will then perfectly fit below the curve. By adding up all these rectangles we get the area below the curve! Let's take a look at an illustration using 70 rectangles.

Forming Riemann sums parabola thin rectangles StudySmarter Approximation of the area below the curve using 70 rectangles - StudySmarter Originals

This seems to fit very well the area below the curve, right? Now imagine what would happen if we used even more rectangles!

Summary

Forming Riemann Sums - Key takeaways

A Riemann sum consists of dividing the area below a curve into rectangles and adding them up.
- Riemann sums are closely related to the left-endpoint and right-endpoint approximations. Both are particular cases of a Riemann sum.
A lower Riemann sum is a Riemann sum obtained by using the least value of each subinterval to calculate the height of each rectangle.
- The value of a lower Riemann sum is always less than or equal to the area below the curve.
An upper Riemann sum is a Riemann sum obtained by using the greatest value of each subinterval to calculate the height of each rectangle.
- The value of an upper Riemann sum is always greater than or equal to the area below the curve.
The area below a curve is bounded between a lower Riemann sum and an upper Riemann sum.
If we take the limit of a Riemann sum as the number of subintervals tends to infinity we get the area below the curve.

Flashcards in Riemann Sum 6

Start learning

All Riemann sums are either a left-endpoint approximation or a right-endpoint approximation.

False.

The right-endpoint approximation is a particular case of a Riemann sum.

True.

An upper Riemann sum is always greater than or equal to the exact area below the curve.

True.

A lower Riemann sum is always less than or equal to the exact area below the curve.

True.

Suppose that, in some interval, a lower Riemann sum of a function is 24.92 and an upper Riemann sum is 25.08. What can we conclude from this information?

That the area below the curve is between 24.92 and 25.08.

A left-endpoint approximation always gives us a lower Riemann sum.

False.

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Frequently Asked Questions about Riemann Sum

What is a Riemann sum?

A Riemann sum consists of dividing the area below a curve into rectangles and adding them up.

How do you find the Riemann sum of a function?

Begin by dividing the interval of the area into subintervals. The more, the better!
Associate a rectangle to each subinterval.
The width of each rectangle is equal to the length of its corresponding subinterval.
The height of each rectangle can be found by evaluating the function at some point within the subinterval.

Usually, the interval is divided regularly. This means that every subinterval will have the same length.

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Riemann Sum

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Sigma notation formulas