Key Concept

#Riemann Sums Study Notes

#Table of Contents

Introduction to Riemann Sums
Types of Riemann Sums
Worked Examples
Practice Questions
Glossary
Summary and Key Takeaways
Exam Strategy

Key Concept

#Introduction to Riemann Sums

A Riemann sum is a technique for approximating the exact value of an accumulation of change, which is often represented as a definite integral or the area between a curve and the x-axis. The basic idea is to sum up the areas of multiple rectangles under a curve to estimate this value.

Key Concept

#Types of Riemann Sums

There are three main types of Riemann sums:

Left Riemann Sum
Right Riemann Sum
Midpoint Riemann Sum

#Left Riemann Sum

#How to 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 $a = x_0 < x_1 < \cdots < x_n = b$ .
Let this define $n$ rectangles.
The width of the $i$ -th rectangle is $(x_i - x_{i-1})$ , and the height is $f(x_{i-1})$ .
The left Riemann sum is the sum of the areas of these $n$ rectangles:

$\sum_{i=1}^{n} (x_i - x_{i-1}) \cdot f(x_{i-1})$

**Example:**

A social sciences researcher is using a function $m$ to model the total mass of garden gnomes appearing on lawns in a neighborhood over time $t$ . The table below gives selected values of $m'(t)$ , the rate of change of the mass, over the time interval $0 \le t \le 12$ . At time $t=0$ , $m(0) = 24.9$ kilograms.

t (days)	0	3	7	10	12
$m'(t)$ (kg/day)	2.6	4.8	12.2	0.7	-1.3

To find the left Riemann sum:

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

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

#Underestimate or Overestimate?

If a function is increasing, the left Riemann sum will be an underestimate.
If a function is decreasing, the left Riemann sum will be an overestimate.

#Right Riemann Sum

#How to 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 $a = x_0 < x_1 < \cdots < x_n = b$ .
Let this define $n$ rectangles.
The width of the $i$ -th rectangle is $(x_i - x_{i-1})$ , and the height is $f(x_i)$ .
The right Riemann sum is the sum of the areas of these $n$ rectangles:

$\sum_{i=1}^{n} (x_i - x_{i-1}) \cdot f(x_i)$

**Example:**

Using the same table of values from the previous example, the right Riemann sum is calculated as follows:

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

The total mass of garden gnomes at $t=12$ is approximately $87.6$ kg after adding the initial mass.

#Underestimate or Overestimate?

If a function is increasing, the right Riemann sum will be an overestimate.
If a function is decreasing, the right Riemann sum will be an underestimate.

#Midpoint Riemann Sum

#How to 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 $a = x_0 < x_1 < \cdots < x_n = b$ .
Let this define $n$ rectangles.
The width of the $i$ -th rectangle is $(x_i - x_{i-1})$ , and the height is $f\left(\frac{x_{i-1} + x_i}{2}\right)$ .
The midpoint Riemann sum is the sum of the areas of these $n$ rectangles:

$\sum_{i=1}^{n} (x_i - x_{i-1}) \cdot f\left(\frac{x_{i-1} + x_i}{2}\right)$

**Example:**

Using the table of values from the previous example, the midpoints for the intervals are 1.5, 5, 8.5, and 11. The midpoint Riemann sum is calculated as follows:

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

The total mass of garden gnomes at $t=12$ is approximately $87.2$ kg after adding the initial mass.

#Underestimate or Overestimate?

It is not immediately obvious whether a midpoint Riemann sum will give an underestimate or an overestimate because the rectangles often combine underestimates and overestimates.

Key Concept

#Worked Examples

#Example 1

A function $f(x)$ is defined on the interval $[0, 4]$ . Using a left Riemann sum with $n=4$ , calculate the sum for $f(x) = x^2$ .

Solution:

Divide the interval $[0, 4]$ into 4 subintervals: $[0, 1]$ , $[1, 2]$ , $[2, 3]$ , $[3, 4]$ .
Calculate the height of each rectangle using the left endpoint:
- $f(0) = 0^2 = 0$
- $f(1) = 1^2 = 1$
- $f(2) = 2^2 = 4$
- $f(3) = 3^2 = 9$
Calculate the left Riemann sum:

$(1-0) \cdot 0 + (2-1) \cdot 1 + (3-2) \cdot 4 + (4-3) \cdot 9 = 0 + 1 + 4 + 9 = 14$

#Example 2

Using the same function $f(x) = x^2$ , calculate the right Riemann sum with $n=4$ .

Solution:

Divide the interval $[0, 4]$ into 4 subintervals: $[0, 1]$ , $[1, 2]$ , $[2, 3]$ , $[3, 4]$ .
Calculate the height of each rectangle using the right endpoint:
- $f(1) = 1^2 = 1$
- $f(2) = 2^2 = 4$
- $f(3) = 3^2 = 9$
- $f(4) = 4^2 = 16$
Calculate the right Riemann sum:

$(1-0) \cdot 1 + (2-1) \cdot 4 + (3-2) \cdot 9 + (4-3) \cdot 16 = 1 + 4 + 9 + 16 = 30$

#Example 3

Using the same function $f(x) = x^2$ , calculate the midpoint Riemann sum with $n=4$ .

Solution:

Divide the interval $[0, 4]$ into 4 subintervals: $[0, 1]$ , $[1, 2]$ , $[2, 3]$ , $[3, 4]$ .
Calculate the height of each rectangle using the midpoint:
- $f(0.5) = (0.5)^2 = 0.25$
- $f(1.5) = (1.5)^2 = 2.25$
- $f(2.5) = (2.5)^2 = 6.25$
- $f(3.5) = (3.5)^2 = 12.25$
Calculate the midpoint Riemann sum:

$(1-0) \cdot 0.25 + (2-1) \cdot 2.25 + (3-2) \cdot 6.25 + (4-3) \cdot 12.25 = 0.25 + 2.25 + 6.25 + 12.25 = 21$

Practice Question

#Practice Questions

Calculate the left Riemann sum for $f(x) = \sin(x)$ on the interval $[0, \pi]$ with $n=4$ .
Calculate the right Riemann sum for $f(x) = e^x$ on the interval $[0, 1]$ with $n=4$ .
Calculate the midpoint Riemann sum for $f(x) = \ln(x+1)$ on the interval $[0, 2]$ with $n=4$ .

Key Concept

#Glossary

Riemann Sum: A method for approximating the value of a definite integral using sums of areas of rectangles.
Definite Integral: The exact accumulation of quantities, often represented as the area under a curve.
Subinterval: A division of the interval over which the function is being integrated.
Left Riemann Sum: Uses the left endpoint of each subinterval to calculate the height of the rectangles.
Right Riemann Sum: Uses the right endpoint of each subinterval to calculate the height of the rectangles.
Midpoint Riemann Sum: Uses the midpoint of each subinterval to calculate the height of the rectangles.

Key Concept

#Summary and Key Takeaways

Riemann sums are a fundamental concept for approximating definite integrals.
Left, right, and midpoint Riemann sums differ in how they choose the height of the rectangles.
Increasing the number of rectangles generally improves the approximation.
Exam tips: Be aware of whether your Riemann sum is an underestimate or an overestimate based on the function's behavior.
Worked examples and practice questions help solidify your understanding.

Exam Tip

#Exam Strategy

Read the question carefully: Identify the type of Riemann sum required.
Divide the interval into subintervals: Ensure you correctly identify the endpoints or midpoints.
Calculate the heights: Use the given function values or calculate them.
Sum the areas: Multiply the heights by the widths and sum them up.
Double-check your calculations: Ensure no arithmetic errors.

By following these strategies and understanding the concepts, you will be well-prepared to tackle Riemann sum problems on your exam. Good luck!