Unit 6.3 – Riemann Sums, Summation Notation, and Definite Integral Notation

AP® Calculus AB & BC | Connecting Summation to Integration

Why This Matters: This topic bridges the gap between discrete sums and continuous integrals! You'll learn summation notation (sigma notation), which is the mathematical shorthand for adding many terms. This leads directly to the formal definition of the definite integral as a limit of Riemann sums. Understanding this connection is crucial—it shows that integration is really just "infinite summation" of infinitely small pieces. This is the theoretical foundation that makes calculus work! Mastering notation and limit properties prepares you for the Fundamental Theorem of Calculus.

∑ Summation Notation (Sigma Notation)

SUMMATION NOTATION BASICS

General Form:

\[ \sum_{i=m}^{n} a_i = a_m + a_{m+1} + a_{m+2} + \cdots + a_n \]

Components:

\(\sum\) = Sigma (Greek letter for "S" = Sum)
\(i\) = index of summation (dummy variable)
\(m\) = lower limit (starting value)
\(n\) = upper limit (ending value)
\(a_i\) = general term (what you're adding)

Examples of Summation Notation:

Example 1: Simple Sum

\[ \sum_{i=1}^{5} i = 1 + 2 + 3 + 4 + 5 = 15 \]

Example 2: Squared Terms

\[ \sum_{k=1}^{4} k^2 = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30 \]

Example 3: Function Evaluation

\[ \sum_{j=0}^{3} (2j + 1) = 1 + 3 + 5 + 7 = 16 \]

📐 Properties of Summation

Essential Summation Properties

1. Constant Multiple Rule:

\[ \sum_{i=m}^{n} c \cdot a_i = c \sum_{i=m}^{n} a_i \]

Constants can be factored out of summation

2. Sum/Difference Rule:

\[ \sum_{i=m}^{n} (a_i \pm b_i) = \sum_{i=m}^{n} a_i \pm \sum_{i=m}^{n} b_i \]

Sum of sums = sum, difference of sums = difference

3. Constant Sum:

\[ \sum_{i=1}^{n} c = nc \]

Adding constant \(c\) exactly \(n\) times

4. Index Shift:

\[ \sum_{i=m}^{n} a_i = \sum_{j=m+k}^{n+k} a_{j-k} \]

Can shift the index (change of variable)

🔢 Special Summation Formulas

🔑 Must-Know Formulas (For AP® Calculus):

Essential Summation Formulas
Sum	Formula	Example (\(n=5\))
\(\sum_{i=1}^{n} 1\)	\(n\)	\(5\)
\(\sum_{i=1}^{n} i\)	\(\frac{n(n+1)}{2}\)	\(\frac{5 \cdot 6}{2} = 15\)
\(\sum_{i=1}^{n} i^2\)	\(\frac{n(n+1)(2n+1)}{6}\)	\(\frac{5 \cdot 6 \cdot 11}{6} = 55\)
\(\sum_{i=1}^{n} i^3\)	\(\left[\frac{n(n+1)}{2}\right]^2\)	\(\left[\frac{5 \cdot 6}{2}\right]^2 = 225\)

💡 Memory Tricks:

\(\sum i\): Think "average times count" = \(\frac{n+1}{2} \cdot n\)
\(\sum i^3\): It's the square of \(\sum i\)!
\(\sum i^2\): Hardest to remember—just memorize the pattern

📊 Riemann Sums Using Summation Notation

Riemann Sums with Sigma Notation

General Riemann Sum:

\[ R_n = \sum_{i=1}^{n} f(x_i^*) \Delta x \]

where \(\Delta x = \frac{b-a}{n}\) and \(x_i^*\) is a sample point

Left Riemann Sum:

\[ L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x = \sum_{i=0}^{n-1} f(a + i\Delta x) \Delta x \]

Right Riemann Sum:

\[ R_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} f(a + i\Delta x) \Delta x \]

Midpoint Riemann Sum:

\[ M_n = \sum_{i=1}^{n} f(m_i) \Delta x = \sum_{i=1}^{n} f\left(a + \left(i-\frac{1}{2}\right)\Delta x\right) \Delta x \]

∫ The Definite Integral: Formal Definition

THE DEFINITE INTEGRAL

Definition as Limit of Riemann Sums:

\[ \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x \]

In words: The definite integral is the limit of Riemann sums as the number of rectangles approaches infinity (and their width approaches zero).

Components of Integral Notation:

\(\int\) = integral sign (elongated S for "sum")
\(a\) = lower limit of integration (starting point)
\(b\) = upper limit of integration (ending point)
\(f(x)\) = integrand (function being integrated)
\(dx\) = differential (indicates variable of integration)

🎯 The Big Connection:

\[ \text{Riemann Sum} \xrightarrow[n \to \infty]{} \text{Definite Integral} \]

\[ \sum_{i=1}^{n} f(x_i^*) \Delta x \xrightarrow[n \to \infty]{} \int_a^b f(x) \, dx \]

As rectangles become infinitely thin and infinitely numerous, the approximation becomes exact!

🔧 Properties of Definite Integrals

Essential Integral Properties

1. Constant Multiple Rule:

\[ \int_a^b c \cdot f(x) \, dx = c \int_a^b f(x) \, dx \]

2. Sum/Difference Rule:

\[ \int_a^b [f(x) \pm g(x)] \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx \]

3. Reversal of Limits:

\[ \int_a^b f(x) \, dx = -\int_b^a f(x) \, dx \]

4. Zero Width:

\[ \int_a^a f(x) \, dx = 0 \]

5. Additivity Over Intervals:

\[ \int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx \]

Can break integral into parts and add them

📖 Comprehensive Worked Examples

Example 1: Evaluating Summation Notation

Problem: Evaluate \(\sum_{k=1}^{5} (3k - 2)\)

Solution:

Method 1: Direct Expansion

\[ \sum_{k=1}^{5} (3k - 2) = (3 \cdot 1 - 2) + (3 \cdot 2 - 2) + (3 \cdot 3 - 2) + (3 \cdot 4 - 2) + (3 \cdot 5 - 2) \]

\[ = 1 + 4 + 7 + 10 + 13 = 35 \]

Method 2: Using Properties

\[ \sum_{k=1}^{5} (3k - 2) = \sum_{k=1}^{5} 3k - \sum_{k=1}^{5} 2 \]

\[ = 3\sum_{k=1}^{5} k - 2(5) \]

\[ = 3 \cdot \frac{5(6)}{2} - 10 = 3(15) - 10 = 45 - 10 = 35 \]

Answer: \(\sum_{k=1}^{5} (3k - 2) = 35\)

Example 2: Simplifying Using Summation Formulas

Problem: Simplify \(\sum_{i=1}^{n} (i^2 + 3i - 5)\)

Solution:

Step 1: Separate the sum

\[ \sum_{i=1}^{n} (i^2 + 3i - 5) = \sum_{i=1}^{n} i^2 + \sum_{i=1}^{n} 3i - \sum_{i=1}^{n} 5 \]

Step 2: Factor out constants

\[ = \sum_{i=1}^{n} i^2 + 3\sum_{i=1}^{n} i - 5n \]

Step 3: Apply formulas

\[ = \frac{n(n+1)(2n+1)}{6} + 3 \cdot \frac{n(n+1)}{2} - 5n \]

\[ = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} - 5n \]

Step 4: Simplify (optional)

Find common denominator (6):

\[ = \frac{n(n+1)(2n+1) + 9n(n+1) - 30n}{6} \]

\[ = \frac{n[(n+1)(2n+1) + 9(n+1) - 30]}{6} \]

Answer: \(\frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} - 5n\)

Example 3: Expressing Riemann Sum in Sigma Notation

Problem: Write the Right Riemann Sum for \(\int_1^5 x^2 \, dx\) with \(n\) subintervals using summation notation.

Solution:

Step 1: Identify components

\(a = 1, b = 5\)
\(\Delta x = \frac{5-1}{n} = \frac{4}{n}\)
For Right Riemann Sum: \(x_i = a + i\Delta x = 1 + i \cdot \frac{4}{n}\)

Step 2: Write summation

\[ R_n = \sum_{i=1}^{n} f(x_i) \Delta x \]

\[ = \sum_{i=1}^{n} \left(1 + \frac{4i}{n}\right)^2 \cdot \frac{4}{n} \]

Step 3: Expand (if needed)

\[ = \sum_{i=1}^{n} \left(1 + \frac{8i}{n} + \frac{16i^2}{n^2}\right) \cdot \frac{4}{n} \]

\[ = \frac{4}{n}\sum_{i=1}^{n} 1 + \frac{32}{n^2}\sum_{i=1}^{n} i + \frac{64}{n^3}\sum_{i=1}^{n} i^2 \]

Answer: \(\sum_{i=1}^{n} \left(1 + \frac{4i}{n}\right)^2 \cdot \frac{4}{n}\)

Example 4: Computing Integral as Limit

Problem: Use the definition of the definite integral to evaluate \(\int_0^2 3x \, dx\).

Solution:

Step 1: Set up Riemann sum

\(a = 0, b = 2, f(x) = 3x\)
\(\Delta x = \frac{2-0}{n} = \frac{2}{n}\)
Use right endpoints: \(x_i = 0 + i \cdot \frac{2}{n} = \frac{2i}{n}\)

Step 2: Write Riemann sum

\[ R_n = \sum_{i=1}^{n} f\left(\frac{2i}{n}\right) \cdot \frac{2}{n} = \sum_{i=1}^{n} 3 \cdot \frac{2i}{n} \cdot \frac{2}{n} \]

\[ = \sum_{i=1}^{n} \frac{12i}{n^2} = \frac{12}{n^2}\sum_{i=1}^{n} i \]

Step 3: Apply summation formula

\[ R_n = \frac{12}{n^2} \cdot \frac{n(n+1)}{2} = \frac{12n(n+1)}{2n^2} = \frac{6(n+1)}{n} = 6\left(1 + \frac{1}{n}\right) \]

Step 4: Take limit

\[ \int_0^2 3x \, dx = \lim_{n \to \infty} R_n = \lim_{n \to \infty} 6\left(1 + \frac{1}{n}\right) = 6(1 + 0) = 6 \]

Answer: \(\int_0^2 3x \, dx = 6\)

💡 Tips, Tricks & Strategies

✅ Summation Notation Tips:

Index is a dummy variable: \(\sum_{i=1}^{n} i = \sum_{k=1}^{n} k = \sum_{j=1}^{n} j\)
Always check limits: Upper and lower bounds determine how many terms
Factor constants first: Makes calculation much easier
Use formulas for \(\sum i, \sum i^2, \sum i^3\): Don't expand large sums
Break complex sums apart: Use sum/difference rule

🔥 Riemann Sum to Integral Tips:

Recognize the pattern: \(\sum f(x_i) \Delta x\) becomes \(\int f(x) \, dx\)
\(\Delta x\) always factors out: Look for \(\frac{1}{n}\) or \(\frac{b-a}{n}\)
As \(n \to \infty\): Terms with \(\frac{1}{n}\) vanish
The limit always exists: For continuous functions on closed intervals
Practice limit properties: \(\lim(1 + \frac{1}{n}) = 1\), \(\lim \frac{c}{n} = 0\)

❌ Common Mistakes to Avoid

Mistake 1: Confusing index limits (starting at 0 vs 1)
Mistake 2: Forgetting to apply summation formulas (expanding huge sums manually)
Mistake 3: Not factoring out constants before summing
Mistake 4: Mixing up \(\sum_{i=1}^{n} i\) and \(\sum_{i=1}^{n} i^2\) formulas
Mistake 5: In Riemann sums, forgetting the \(\Delta x\) factor
Mistake 6: Incorrectly computing \(\Delta x = \frac{b-a}{n}\)
Mistake 7: For right endpoints, using \(x_i = a + (i-1)\Delta x\) (that's left!)
Mistake 8: Not simplifying before taking limits
Mistake 9: Forgetting that \(\lim_{n \to \infty} \frac{1}{n} = 0\)
Mistake 10: Treating \(\int\) as a finite sum (it's an infinite limit!)

📝 Practice Problems

Set A: Summation Notation

Evaluate: \(\sum_{k=1}^{6} (2k + 3)\)
Simplify: \(\sum_{i=1}^{n} (i^2 - i)\)
Write in expanded form: \(\sum_{j=0}^{4} j^3\)

Answers:

\(2 \cdot \frac{6 \cdot 7}{2} + 3(6) = 42 + 18 = 60\)
\(\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2}\)
\(0 + 1 + 8 + 27 + 64 = 100\)

Set B: Riemann Sums with Sigma

Express Left Riemann Sum for \(\int_0^4 (x+1) \, dx\) with \(n\) subintervals
Evaluate: \(\lim_{n \to \infty} \sum_{i=1}^{n} \frac{2i}{n^2}\)

Answers:

\(\sum_{i=0}^{n-1} \left(\frac{4i}{n} + 1\right) \cdot \frac{4}{n}\)
\(\frac{2}{n^2} \cdot \frac{n(n+1)}{2} = \frac{n+1}{n} \to 1\) as \(n \to \infty\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Proper notation: Use \(\sum\) and \(\int\) correctly
Clear setup: Show \(\Delta x\) and sample points
Summation formulas: State which formula you're using
Limit notation: Write \(\lim_{n \to \infty}\) explicitly
Simplification shown: Don't skip algebraic steps
Units (if applicable): Include in context problems

⚡ Ultimate Quick Reference

ESSENTIAL FORMULAS SUMMARY

Must-Know for AP® Calculus
Category	Formula
\(\Delta x\)	\(\frac{b-a}{n}\)
\(\sum_{i=1}^{n} 1\)	\(n\)
\(\sum_{i=1}^{n} i\)	\(\frac{n(n+1)}{2}\)
\(\sum_{i=1}^{n} i^2\)	\(\frac{n(n+1)(2n+1)}{6}\)
\(\sum_{i=1}^{n} i^3\)	\(\left[\frac{n(n+1)}{2}\right]^2\)
Riemann Sum	\(\sum_{i=1}^{n} f(x_i^*) \Delta x\)
Definite Integral	\(\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x\)

Master the Notation! Summation notation (\(\sum\)) is mathematical shorthand for adding many terms. The key formulas are: \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\), \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\), and \(\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2\). Riemann sums can be written compactly as \(\sum_{i=1}^{n} f(x_i^*) \Delta x\), and the definite integral is the limit of these sums: \(\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x\). This shows integration is "infinite summation"—as rectangles become infinitely thin (\(\Delta x \to 0\)) and infinitely numerous (\(n \to \infty\)), approximation becomes exact. Properties of sums transfer to integrals: constant multiple rule, sum/difference rule, and additivity. Always remember \(\Delta x = \frac{b-a}{n}\), factor out constants, and use summation formulas instead of expanding. The notation \(\int_a^b f(x) \, dx\) represents the signed area under the curve from \(a\) to \(b\). This formal foundation prepares you for the Fundamental Theorem of Calculus! 🎯✨