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Riemann Sums & Definite Integrals

David Brown

David Brown

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Study Guide Overview

This study guide covers Riemann Sums as a method for approximating definite integrals. It explains the three main types: Left, Right, and Midpoint Riemann Sums, including how to calculate them and whether they overestimate or underestimate the integral. The guide provides worked examples, practice questions, a glossary of terms, and exam strategies.

Key Concept

#Riemann Sums Study Notes

#Table of Contents

  1. Introduction to Riemann Sums
  2. Types of Riemann Sums
    • Left Riemann Sum
    • Right Riemann Sum
    • Midpoint Riemann Sum
  3. Worked Examples
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways
  7. 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 fff between x=ax=ax=a and x=bx=bx=b (where a<ba < ba<b):

  1. Divide the interval into nnn subintervals by choosing values x0,x1,...,xnx_0, x_1, ..., x_nx0​,x1​,...,xn​ such that a=x0<x1<⋯<xn=ba = x_0 < x_1 < \cdots < x_n = ba=x0​<x1​<⋯<xn​=b.
  2. Let this define nnn rectangles.
  3. The width of the iii-th rectangle is (xi−xi−1)(x_i - x_{i-1})(xi​−xi−1​), and the height is f(xi−1)f(x_{i-1})f(xi−1​).
  4. The left Riemann sum is the sum of the areas of these nnn rectangles:

∑i=1n(xi−xi−1)⋅f(xi−1)\sum_{i=1}^{n} (x_i - x_{i-1}) \cdot f(x_{i-1})i=1∑n​(xi​−xi−1​)⋅f(xi−1​)

**Example:**

A social sciences researcher is using a function mmm to model the total mass of garden gnomes appearing on lawns in a neighborhood over time ttt. The table below gives selected values of m′(t)m'(t)m′(t), the rate of change of the mass, over the time interval 0≤t≤120 \le t \le 120≤t≤12. At time t=0t=0t=0, m(0)=24.9m(0) = 24.9m(0)=24.9 kilograms.

t (days)0371012
m′(t)m'(t)m′(t) (kg/day)2.64.812.20.7-1.3

To find the left Riemann sum:

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

The mass of garden gnomes increases by approximately 65.0 kg between t=0t=0t=0 and t=12t=12t=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 fff between x=ax=ax=a and x=bx=bx=b (where a<ba < ba<b):

  1. Divide the interval into nnn subintervals by choosing values x0,x1,...,xnx_0, x_1, ..., x_nx0​,x1​,...,xn​ such that a=x0<x1<⋯<xn=ba = x_0 < x_1 < \cdots < x_n = ba=x0​<x1​<⋯<xn​=b.
  2. Let this define nnn rectangles.
  3. The width of the iii-th rectangle is (xi−xi−1)(x_i - x_{i-1})(xi​−xi−1​), and the height is f(xi)f(x_i)f(xi​).
  4. The right Riemann sum is the sum of the areas of these nnn rectangles:

∑i=1n(xi−xi−1)⋅f(xi)\sum_{i=1}^{n} (x_i - x_{i-1}) \cdot f(x_i)i=1∑n​(xi​−xi−1​)⋅f(xi​)

**Example:**

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

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

The total mass of garden gnomes at t=12t=12t=12 is approximately 87.687.687.6 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 fff between x=ax=ax=a and x=bx=bx=b (where a<ba < ba<b):

  1. Divide the interval into nnn subintervals by choosing values x0,x1,...,xnx_0, x_1, ..., x_nx0​,x1​,...,xn​ such that a=x0<x1<⋯<xn=ba = x_0 < x_1 < \cdots < x_n = ba=x0​<x1​<⋯<xn​=b.
  2. Let this define nnn rectangles.
  3. The width of the iii-th rectangle is (xi−xi−1)(x_i - x_{i-1})(xi​−xi−1​), and the height is f(xi−1+xi2)f\left(\frac{x_{i-1} + x_i}{2}\right)f(2xi−1​+xi​​).
  4. The midpoint Riemann sum is the sum of the areas of these nnn rectangles:

∑i=1n(xi−xi−1)⋅f(xi−1+xi2)\sum_{i=1}^{n} (x_i - x_{i-1}) \cdot f\left(\frac{x_{i-1} + x_i}{2}\right)i=1∑n​(xi​−xi−1​)⋅f(2xi−1​+xi​​)

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

The total mass of garden gnomes at t=12t=12t=12 is approximately 87.287.287.2 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)f(x)f(x) is defined on the interval [0,4][0, 4][0,4]. Using a left Riemann sum with n=4n=4n=4, calculate the sum for f(x)=x2f(x) = x^2f(x)=x2.

Solution:

  1. Divide the interval [0,4][0, 4][0,4] into 4 subintervals: [0,1][0, 1][0,1], [1,2][1, 2][1,2], [2,3][2, 3][2,3], [3,4][3, 4][3,4].
  2. Calculate the height of each rectangle using the left endpoint:
    • f(0)=02=0f(0) = 0^2 = 0f(0)=02=0
    • f(1)=12=1f(1) = 1^2 = 1f(1)=12=1
    • f(2)=22=4f(2) = 2^2 = 4f(2)=22=4
    • f(3)=32=9f(3) = 3^2 = 9f(3)=32=9
  3. Calculate the left Riemann sum:

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

#Example 2

Using the same function f(x)=x2f(x) = x^2f(x)=x2, calculate the right Riemann sum with n=4n=4n=4.

Solution:

  1. Divide the interval [0,4][0, 4][0,4] into 4 subintervals: [0,1][0, 1][0,1], [1,2][1, 2][1,2], [2,3][2, 3][2,3], [3,4][3, 4][3,4].
  2. Calculate the height of each rectangle using the right endpoint:
    • f(1)=12=1f(1) = 1^2 = 1f(1)=12=1
    • f(2)=22=4f(2) = 2^2 = 4f(2)=22=4
    • f(3)=32=9f(3) = 3^2 = 9f(3)=32=9
    • f(4)=42=16f(4) = 4^2 = 16f(4)=42=16
  3. Calculate the right Riemann sum:

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

#Example 3

Using the same function f(x)=x2f(x) = x^2f(x)=x2, calculate the midpoint Riemann sum with n=4n=4n=4.

Solution:

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

(1−0)⋅0.25+(2−1)⋅2.25+(3−2)⋅6.25+(4−3)⋅12.25=0.25+2.25+6.25+12.25=21(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(1−0)⋅0.25+(2−1)⋅2.25+(3−2)⋅6.25+(4−3)⋅12.25=0.25+2.25+6.25+12.25=21

Practice Question

#Practice Questions

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

  1. Read the question carefully: Identify the type of Riemann sum required.
  2. Divide the interval into subintervals: Ensure you correctly identify the endpoints or midpoints.
  3. Calculate the heights: Use the given function values or calculate them.
  4. Sum the areas: Multiply the heights by the widths and sum them up.
  5. 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!

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Question 1 of 12

What is the primary purpose of using a Riemann sum? 🤔

To find the exact area of a rectangle

To approximate the area under a curve

To calculate the derivative of a function

To find the x-intercept of a function