Riemann Sums (College Board AP® Calculus AB)

Revision Note

Author

Roger B

Expertise

Maths

What is a Riemann sum?

A Riemann sum is a method for approximating the exact value of an accumulation of change
- Equivalently, it is a method for approximating the exact value of a definite integral
- Or the exact area between a curve and the x-axis
The approximation is made by adding up the areas of a number of rectangles
There are three types of Riemann sum you should know and be able to use:
- Left Riemann sum
- Right Riemann sum
- Midpoint Riemann sum

Left Riemann sum

How do I calculate a left Riemann sum?

To calculate the left Riemann sum of a function $f$ between $x = a$ and $x = b$ (where $a < b$ ):
- Divide the interval into $n$ subintervals by choosing values $x_{0}, x_{1}, . . ., x_{n}$ such that $a = x_{0} < x_{1} < \dots < x_{n} = b$
  - The intervals do not need to be the same size
- Let this define $n$ rectangles
  - The $i$ th rectangle has a width of $(x_{i} - x_{i - 1})$
    - This is the distance from the left-hand side of the rectangle to the right-hand side
  - The $i$ th rectangle has a height of $f (x_{i - 1})$
    - This is the value of the function at the left-hand side of the rectangle
- The left Riemann sum is the sum of the areas of these $n$ rectangles
  - $(x_{1} - x_{0}) \cdot f (x_{0}) + (x_{2} - x_{1}) \cdot f (x_{1}) + \dots + (x_{n} - x_{n - 1}) \cdot f (x_{n - 1})$

**An example of the graph of a curve, showing the rectangles used to calculate a left Riemann sum**

In general, increasing the number of rectangles, $n$ , gives a more accurate approximation
On the exam you may just be given values of the function in a table, rather than being given the function explicitly
- See the Worked Example

How can I tell if a left Riemann sum is an underestimate or an overestimate?

If a function is increasing over the interval for which a left Riemann sum is being calculated
- then the left Riemann sum will be an underestimate
If a function is decreasing over the interval for which a left Riemann sum is being calculated
- then the left Riemann sum will be an overestimate

Two graphs with the rectangles used to calculate a left Riemann sum, showing that a left Riemann sum will give an underestimate for an increasing function and an overestimate for a decreasing function

If a function has both increasing and decreasing portions, then it is not immediately obvious whether a Riemann sum will be an underestimate or an overestimate

Worked Example

A social sciences researcher is using a function $m$ to model the total mass of all the garden gnomes appearing on lawns in a particular neighborhood at time $t$ . The function $m$ is twice-differentiable, with $m (t)$ measured in kilograms and $t$ measured in days.

The table below gives selected values of $m^{'} (t)$ , the rate of change of the mass, over the time interval $0 \leq t \leq 12$ . At time $t = 0$ , $m (0) = 24.9$ kilograms.

$t$

(days)

$m^{'} (t)$

(kilograms per day)

2.6

4.8

12.2

0.7

-1.3

(a) Use a left Riemann sum with the four subintervals indicated in the table to estimate the change in the total mass of garden gnomes between $t = 0$ and $t = 12$ .

Answer:

The Riemann sum will be based on four rectangles

The first rectangle will have a width of (3-0) and a height equal to $m (0)$
The second rectangle will have a width of (7-3) and a height equal to $m (3)$
The third rectangle will have a width of (10-7) and a height equal to $m (7)$
The fourth rectangle will have a width of (12-10) and a height equal to $m (10)$

$(3 - 0) \cdot 2.6 + (7 - 3) \cdot 4.8 + (10 - 7) \cdot 12.2 + (12 - 10) \cdot 0.7 = 65.0$

This has units of kilograms, because each term is the product of a 'kg/day' quantity and a 'days' quantity

The mass of garden gnomes increases by approximately 65.0 kg between $t = 0$ and $t = 12$

(b) Hence find an estimate for the total mass of the garden gnomes at $t = 12$ .

Answer:

The answer to part (a) only shows the change in mass

To find the estimate for the total final mass, add the initial mass of 24.9 kilograms

$65.0 + 24.9 = 89.9$

89.9 kg

Right Riemann sum

How do I calculate a right Riemann sum?

To calculate the right Riemann sum of a function $f$ between $x = a$ and $x = b$ (where $a < b$ ):
- Divide the interval into $n$ subintervals by choosing values $x_{0}, x_{1}, . . ., x_{n}$ such that $a = x_{0} < x_{1} < \dots < x_{n} = b$
  - The intervals do not need to be the same size
- Let this define $n$ rectangles
  - The $i$ th rectangle has a width of $(x_{i} - x_{i - 1})$
    - This is the distance from the left-hand side of the rectangle to the right-hand side
  - The $i$ th rectangle has a height of $f (x_{i})$
    - This is the value of the function at the right-hand side of the rectangle
- The right Riemann sum is the sum of the areas of these $n$ rectangles
  - $(x_{1} - x_{0}) \cdot f (x_{1}) + (x_{2} - x_{1}) \cdot f (x_{2}) + \dots + (x_{n} - x_{n - 1}) \cdot f (x_{n})$

An example of the graph of a curve, showing the rectangles used to calculate a right Riemann sum with n=5 — **An example of a right Riemann sum with n=5**

In general, increasing the number of rectangles, $n$ , gives a more accurate approximation
On the exam you may just be given values of the function in a table, rather than being given the function explicitly
- See the Worked Example

How can I tell if a right Riemann sum is an underestimate or an overestimate?

If a function is increasing over the interval for which a right Riemann sum is being calculated
- then the right Riemann sum will be an overestimate
If a function is decreasing over the interval for which a right Riemann sum is being calculated
- then the right Riemann sum will be an underestimate

Two graphs with the rectangles used to calculate a right Riemann sum, showing that a right Riemann sum will give an overestimate for an increasing function and an underestimate for a decreasing function

If a function has both increasing and decreasing portions, then it is not immediately obvious whether a Riemann sum will be an underestimate or an overestimate

Worked Example

The table below gives selected values of $m^{'} (t)$ , the rate of change of the mass, over the time interval $0 \leq t \leq 12$ . At time $t = 0$ , $m (0) = 24.9$ kilograms.

$t$

(days)

$m^{'} (t)$

(kilograms per day)

2.6

4.8

12.2

0.7

-1.3

Use a right Riemann sum with the four subintervals indicated in the table to find an estimate for the total mass of the garden gnomes at $t = 12$ .

Answer:

The Riemann sum will be based on four rectangles

The first rectangle will have a width of (3-0) and a height equal to $m (3)$
The second rectangle will have a width of (7-3) and a height equal to $m (7)$
The third rectangle will have a width of (10-7) and a height equal to $m (10)$
The fourth rectangle will have a width of (12-10) and a height equal to $m (12)$

$(3 - 0) \cdot 4.8 + (7 - 3) \cdot 12.2 + (10 - 7) \cdot 0.7 + (12 - 10) \cdot (- 1.3) = 62.7$

This has units of kilograms, because each term is the product of a 'kg/day' quantity and a 'days' quantity

However that answer only approximates the change in mass

To find the estimate for the total final mass, add the initial mass of 24.9 kilograms

$62.7 + 24.9 = 87.6$

The total mass of garden gnomes at $t = 12$ is approximately 87.6 kg

Midpoint Riemann sum

How do I calculate a midpoint Riemann sum?

To calculate the midpoint Riemann sum of a function $f$ between $x = a$ and $x = b$ (where $a < b$ ):
- Divide the interval into $n$ subintervals by choosing values $x_{0}, x_{1}, . . ., x_{n}$ such that $a = x_{0} < x_{1} < \dots < x_{n} = b$
  - The intervals do not need to be the same size
- Let this define $n$ rectangles
  - The $i$ th rectangle has a width of $(x_{i} - x_{i - 1})$
    - This is the distance from the left-hand side of the rectangle to the right-hand side
  - The $i$ th rectangle has a height of $f (\frac{x_{i - 1} + x_{i}}{2})$
    - This is the value of the function at the midpoint of the rectangle
- The right Riemann sum is the sum of the areas of these $n$ rectangles
  - $(x_{1} - x_{0}) \cdot f (\frac{x_{0} + x_{1}}{2}) + (x_{2} - x_{1}) \cdot f (\frac{x_{1} + x_{2}}{2}) + \dots + (x_{n} - x_{n - 1}) \cdot f (\frac{x_{n - 1} + x_{n}}{2})$

An example of the graph of a curve, showing the rectangles used to calculate a midpoint Riemann sum with n=5 — **An example of a midpoint Riemann sum with n=5**

It is not immediately obvious whether a midpoint Riemann sum will give an underestimate or an overestimate
- The rectangles often combine an underestimate on one side of their midpoint line, and an overestimate on the other side
In general, increasing the number of rectangles, $n$ , gives a more accurate approximation
On the exam you may just be given values of the function in a table, rather than being given the function explicitly
- See the Worked Example

Worked Example

The table below gives selected values of $m^{'} (t)$ , the rate of change of the mass, over the time interval $0 \leq t \leq 12$ . At time $t = 0$ , $m (0) = 24.9$ kilograms.

$t$

(days)

1.5

8.5

$m^{'} (t)$

(kg/ day)

2.6

3.5

4.8

9.1

12.2

5.2

0.7

-0.1

-1.3

Use a midpoint Riemann sum to find an estimate for the total mass of the garden gnomes at $t = 12$ . Use four subintervals to calculate the sum, for subintervals of $t$ between 0 and 3, 3 and 7, 7 and 10, and 10 and 12.

Answer:

Note that 1.5, 5, 8.5 and 11 are the midpoints of the intervals 0-3, 3-7, 7-10 and 10-12, respectively

The Riemann sum will be based on four rectangles

The first rectangle will have a width of (3-0) and a height equal to $m (1.5)$
The second rectangle will have a width of (7-3) and a height equal to $m (5)$
The third rectangle will have a width of (10-7) and a height equal to $m (8.5)$
The fourth rectangle will have a width of (12-10) and a height equal to $m (11)$

$(3 - 0) \cdot 3.5 + (7 - 3) \cdot 9.1 + (10 - 7) \cdot 5.2 + (12 - 10) \cdot (- 0.1) = 62.3$

This has units of kilograms, because each term is the product of a 'kg/day' quantity and a 'days' quantity

However that answer only approximates the change in mass

To find the estimate for the total final mass, add the initial mass of 24.9 kilograms

$62.3 + 24.9 = 87.2$

The total mass of garden gnomes at $t = 12$ is approximately 87.2 kg

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Riemann Sums (College Board AP® Calculus AB)

Revision Note

What is a Riemann sum?

Left Riemann sum

How do I calculate a left Riemann sum?

How can I tell if a left Riemann sum is an underestimate or an overestimate?

Worked Example

Right Riemann sum

How do I calculate a right Riemann sum?

How can I tell if a right Riemann sum is an underestimate or an overestimate?

Worked Example

Midpoint Riemann sum

How do I calculate a midpoint Riemann sum?

Worked Example

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Author: Roger B