# Left & right Riemann sums

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.

### 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 |
---|---|

## Riemann sum problems with graphs

Imagine we're asked to approximate the area between $y=g(x)$ and the $x$-axis from $x=2$ to $x=6$.

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.

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

Imagine we're asked to approximate the area between the $x$-axis and the graph of $f$ from $x = 1$ to $x = 10$ 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 $9\div 3=\blueD3$.

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 $f(4)=\goldD{8}$.In a similar way we can find that the second rectangle, that sits on the interval $[4,7]$, has its top-right vertex at $f(7)=\purpleD{3}$.

Our third (and last) rectangle has its top-right vertex at $f(10)=\maroonD{5}$.

Now all that remains is to crunch the numbers.

First rectangle | Second rectangle | Third rectangle | |
---|---|---|---|

Width | $\blueD{3}$ | $\blueD{3}$ | $\blueD{3}$ |

Height | $\goldD{8}$ | $\purpleD{3}$ | $\maroonD{5}$ |

Area | $\blueD{3}\cdot\goldD{8}=24$ | $\blueD{3}\cdot\purpleD{3}=9$ | $\blueD{3}\cdot\maroonD{5}=15$ |

Then, after finding the individual areas, we'd add them up to get our approximation: $48$ units$^2$.

Now imagine we're asked to approximate the area between the $x$-axis and the graph of $f(x) = 2^x$ from $x =-3$ to $x = 3$ using a right Riemann sum with three equal subdivisions.

The entire interval $[-3,3]$ is $6$ units wide, so each of the three rectangles should be $6 \div 3 = \blueD{2}$ units wide.

The first rectangle sits on $[-3,-1]$, so its height is $f(-1)=2^{-1}=\goldD{0.5}$. Similarly, the height of the second rectangle is $f(1)=2^1=\purpleD{2}$ and the height of the third rectangle is $f(3)=2^3=\maroonD{8}$.

First rectangle | Second rectangle | Third rectangle | |
---|---|---|---|

Width | $\blueD{2}$ | $\blueD{2}$ | $\blueD{2}$ |

Height | $\goldD{0.5}$ | $\purpleD{2}$ | $\maroonD{8}$ |

Area | $\blueD{2}\cdot\goldD{0.5}=1$ | $\blueD{2}\cdot\purpleD{2}=4$ | $\blueD{2}\cdot\maroonD{8}=16$ |

So our approximation is $21$ 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).

*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$\quad$ | Right Riemann sum |
---|---|

#### 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 |