Unit 6.6 – Applying Properties of Definite Integrals

AP® Calculus AB & BC | Essential Properties and Applications

Why This Matters: The properties of definite integrals are powerful tools that simplify calculations and allow you to manipulate integrals algebraically. These properties—linearity, additivity, comparison theorems—let you split complex integrals, combine simpler ones, and make estimates without explicit calculation. The average value of a function and the Mean Value Theorem for Integrals are also critical concepts. Mastering these properties is essential for both computational efficiency and theoretical understanding. They appear frequently on AP® exams!

🎯 Core Properties of Definite Integrals

The Essential Properties

Property 1: Constant Multiple Rule

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

Constants factor out of integrals

Property 2: Sum/Difference Rule

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

Integral of sum = sum of integrals

Property 3: Reversal of Limits

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

Swapping limits changes sign

Property 4: Zero Width

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

Integral from point to itself is zero

Property 5: Additivity Over Intervals

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

Can split integral at any interior point

🔑 Linearity Property (Combining 1 and 2):

\[ \int_a^b [c_1f(x) + c_2g(x)] \, dx = c_1\int_a^b f(x)\,dx + c_2\int_a^b g(x)\,dx \]

This is THE most used property—integrals are linear operators!

📊 Comparison Properties

Inequality Properties

Property 6: Preserves Inequalities

\[ \text{If } f(x) \leq g(x) \text{ on } [a,b], \text{ then } \int_a^b f(x)\,dx \leq \int_a^b g(x)\,dx \]

Larger function → larger integral (when \(a < b\))

Property 7: Non-negative Function

\[ \text{If } f(x) \geq 0 \text{ on } [a,b], \text{ then } \int_a^b f(x)\,dx \geq 0 \]

Positive function → positive integral

Property 8: Bounds on Integrals

\[ \text{If } m \leq f(x) \leq M \text{ on } [a,b], \text{ then } m(b-a) \leq \int_a^b f(x)\,dx \leq M(b-a) \]

Bounded function → bounded integral

📈 Average Value of a Function

AVERAGE VALUE FORMULA

The Average Value of \(f\) on \([a,b]\):

\[ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx \]

Interpretation: The constant height that gives the same area as \(f(x)\)

Understanding Average Value:

Geometric meaning: Average height of the function over the interval
Formula breakdown: Total area ÷ width of interval
Units: Same as \(f(x)\) (not area units!)
Comparison: Similar to arithmetic mean: sum ÷ count

Area Interpretation:

\[ \int_a^b f(x)\,dx = f_{\text{avg}} \cdot (b-a) \]

Rectangle with height \(f_{\text{avg}}\) has same area as region under \(f\)

⭐ Mean Value Theorem for Integrals

MVT for Integrals

THE STATEMENT:

\[ \text{If } f \text{ is continuous on } [a,b], \text{ then } \exists c \in (a,b) \text{ such that:} \]

\[ f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx = f_{\text{avg}} \]

In words: There exists at least one point where the function equals its average value

Alternative Form:

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

for some \(c\) in \([a,b]\)

📝 Important Notes:

Requires continuity: Function must be continuous on \([a,b]\)
\(c\) exists but may not be unique: Could be multiple values
Geometric interpretation: Function crosses its average value at \(c\)
Cannot always find \(c\) exactly: Existence theorem, not constructive

📖 Comprehensive Worked Examples

Example 1: Using Linearity Property

Problem: Given \(\int_0^3 f(x)\,dx = 5\) and \(\int_0^3 g(x)\,dx = -2\), find:
(a) \(\int_0^3 [2f(x) + 3g(x)]\,dx\)
(b) \(\int_0^3 [f(x) - 4g(x)]\,dx\)

Solution:

Part (a):

\[ \int_0^3 [2f(x) + 3g(x)]\,dx = 2\int_0^3 f(x)\,dx + 3\int_0^3 g(x)\,dx \]

\[ = 2(5) + 3(-2) = 10 - 6 = 4 \]

Part (b):

\[ \int_0^3 [f(x) - 4g(x)]\,dx = \int_0^3 f(x)\,dx - 4\int_0^3 g(x)\,dx \]

\[ = 5 - 4(-2) = 5 + 8 = 13 \]

Answers: (a) 4 | (b) 13

Example 2: Additivity Over Intervals

Problem: Given \(\int_1^5 f(x)\,dx = 12\) and \(\int_1^3 f(x)\,dx = 7\), find \(\int_3^5 f(x)\,dx\).

Solution:

Apply additivity property:

\[ \int_1^5 f(x)\,dx = \int_1^3 f(x)\,dx + \int_3^5 f(x)\,dx \]

\[ 12 = 7 + \int_3^5 f(x)\,dx \]

\[ \int_3^5 f(x)\,dx = 12 - 7 = 5 \]

Answer: \(\int_3^5 f(x)\,dx = 5\)

Example 3: Average Value

Problem: Find the average value of \(f(x) = x^2 + 1\) on the interval \([0, 3]\).

Solution:

Step 1: Apply average value formula

\[ f_{\text{avg}} = \frac{1}{3-0}\int_0^3 (x^2 + 1)\,dx = \frac{1}{3}\int_0^3 (x^2 + 1)\,dx \]

Step 2: Evaluate the integral

\[ \int_0^3 (x^2 + 1)\,dx = \left[\frac{x^3}{3} + x\right]_0^3 = \left(\frac{27}{3} + 3\right) - 0 = 9 + 3 = 12 \]

Step 3: Complete the calculation

\[ f_{\text{avg}} = \frac{1}{3}(12) = 4 \]

Answer: The average value is 4

Example 4: Using Comparison Property

Problem: Without calculating, determine which integral is larger:
\(A = \int_0^1 x^2\,dx\) or \(B = \int_0^1 x^3\,dx\)

Solution:

Compare the functions on \([0,1]\):

For \(0 \leq x \leq 1\):

\[ x^3 = x \cdot x^2 \leq x^2 \quad \text{(since } x \leq 1\text{)} \]

Therefore: \(x^3 \leq x^2\) on \([0,1]\)

Apply comparison property:

Since \(x^3 \leq x^2\) on \([0,1]\):

\[ \int_0^1 x^3\,dx \leq \int_0^1 x^2\,dx \]

Therefore: \(B \leq A\), so \(A\) is larger

Verification (optional):

\(A = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}\)
\(B = \left[\frac{x^4}{4}\right]_0^1 = \frac{1}{4}\)
Indeed: \(\frac{1}{4} < \frac{1}{3}\) ✓

Answer: \(\int_0^1 x^2\,dx\) is larger

Example 5: Mean Value Theorem for Integrals

Problem: Verify the Mean Value Theorem for Integrals for \(f(x) = x^2\) on \([0, 2]\). Find the value(s) of \(c\).

Solution:

Step 1: Calculate average value

\[ f_{\text{avg}} = \frac{1}{2-0}\int_0^2 x^2\,dx = \frac{1}{2}\left[\frac{x^3}{3}\right]_0^2 = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3} \]

Step 2: Find \(c\) where \(f(c) = f_{\text{avg}}\)

\[ c^2 = \frac{4}{3} \]

\[ c = \pm\sqrt{\frac{4}{3}} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3} \]

Step 3: Check which value is in \([0,2]\)

Since \(c\) must be in \([0, 2]\), we take the positive value:

\[ c = \frac{2\sqrt{3}}{3} \approx 1.155 \]

This is in \((0, 2)\) ✓

Answer: \(c = \frac{2\sqrt{3}}{3}\) (approximately 1.155)

📊 Special Case: Absolute Value

Absolute Value Property:

Triangle Inequality for Integrals:

\[ \left|\int_a^b f(x)\,dx\right| \leq \int_a^b |f(x)|\,dx \]

Absolute value of integral ≤ integral of absolute value

Interpretation:

Left side: Net signed area (cancellation occurs)
Right side: Total area (no cancellation)
Example: For velocity, left = displacement, right = distance

💡 Essential Tips & Strategies

✅ Property Application Tips:

Linearity first: Always factor out constants and split sums/differences
Split at discontinuities: Use additivity if function has breaks
Estimate before computing: Use bounds to check reasonableness
Average value units: Same as function, not area!
MVT existence: Guarantees \(c\) exists, but may not find it explicitly
Comparison for inequalities: Don't need exact values

🎯 Common Problem Types:

Quick Strategy Guide
If Problem Asks...	Use This Property
Combine multiple integrals	Linearity (sum/difference + constant multiple)
Split an integral	Additivity over intervals
Compare integrals without calculating	Comparison property
Find average/mean value	Average value formula
Show \(f(c)\) equals average	Mean Value Theorem for Integrals
Estimate bounds on integral	\(m(b-a) \leq \int_a^b f \leq M(b-a)\)

❌ Common Mistakes to Avoid

Mistake 1: Thinking \(\int_a^b [f(x) \cdot g(x)]\,dx = \int_a^b f(x)\,dx \cdot \int_a^b g(x)\,dx\) (FALSE! No product rule for integrals)
Mistake 2: Confusing average value with average rate of change
Mistake 3: Using wrong units for average value (should match function, not area)
Mistake 4: Not checking if \(c\) from MVT is in the interval \((a,b)\)
Mistake 5: Forgetting negative sign when reversing limits
Mistake 6: Applying comparison property when limits are reversed (\(b < a\))
Mistake 7: Thinking linearity applies to products or quotients
Mistake 8: Not simplifying before integrating (miss opportunity to use properties)
Mistake 9: Confusing \(\left|\int f\right|\) with \(\int |f|\)
Mistake 10: Assuming MVT gives unique \(c\) (may be multiple values)

📝 Practice Problems

Set A: Properties

If \(\int_2^7 f(x)\,dx = 10\), find \(\int_7^2 f(x)\,dx\)
Given \(\int_0^4 f(x)\,dx = 6\) and \(\int_0^4 g(x)\,dx = -2\), find \(\int_0^4 [3f(x) - 2g(x)]\,dx\)
If \(\int_1^5 f(x)\,dx = 8\) and \(\int_3^5 f(x)\,dx = 3\), find \(\int_1^3 f(x)\,dx\)

Answers:

\(-10\) (reverse limits → negative)
\(3(6) - 2(-2) = 18 + 4 = 22\)
\(8 - 3 = 5\)

Set B: Average Value

Find the average value of \(f(x) = 3x^2\) on \([0, 2]\)
The average value of \(f\) on \([1, 5]\) is 7. Find \(\int_1^5 f(x)\,dx\)

Answers:

\(\frac{1}{2}\int_0^2 3x^2\,dx = \frac{1}{2}[x^3]_0^2 = \frac{1}{2}(8) = 4\)
\(f_{\text{avg}} = \frac{1}{4}\int_1^5 f(x)\,dx = 7\), so \(\int_1^5 f(x)\,dx = 28\)

Set C: Applications

If \(2 \leq f(x) \leq 5\) on \([0, 4]\), what are the bounds on \(\int_0^4 f(x)\,dx\)?

Answer:

\(2(4) \leq \int_0^4 f(x)\,dx \leq 5(4)\), so \(8 \leq \int_0^4 f(x)\,dx \leq 20\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Property identification: State which property you're using
Clear algebraic steps: Show how you apply linearity
Correct formula: For average value, show \(\frac{1}{b-a}\int_a^b f(x)\,dx\)
Units included: Especially for average value
Justification: For comparison, explain why one function is larger
Verification: Check if \(c\) is in the interval for MVT
Complete reasoning: Don't just state "by properties"—show which one!

⚡ Ultimate Quick Reference

ESSENTIAL PROPERTIES SUMMARY

Complete Property Reference
Property	Formula
Constant Multiple	\(\int_a^b c \cdot f = c\int_a^b f\)
Sum/Difference	\(\int_a^b (f \pm g) = \int_a^b f \pm \int_a^b g\)
Reverse Limits	\(\int_a^b f = -\int_b^a f\)
Zero Width	\(\int_a^a f = 0\)
Additivity	\(\int_a^b f + \int_b^c f = \int_a^c f\)
Comparison	If \(f \leq g\), then \(\int_a^b f \leq \int_a^b g\)
Bounds	If \(m \leq f \leq M\), then \(m(b-a) \leq \int_a^b f \leq M(b-a)\)
Average Value	\(f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx\)
MVT for Integrals	\(\exists c: f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx\)

Master the Properties! The properties of definite integrals are algebraic tools that make integration practical. Linearity (constant multiple + sum/difference rules) is most used—always factor out constants and split sums. Additivity lets you split integrals at any point: \(\int_a^b + \int_b^c = \int_a^c\). Reversal of limits changes sign: \(\int_a^b = -\int_b^a\). Comparison properties let you estimate without calculating: if \(f \leq g\), then \(\int f \leq \int g\), and if \(m \leq f \leq M\), then \(m(b-a) \leq \int_a^b f \leq M(b-a)\). The average value of \(f\) on \([a,b]\) is \(\frac{1}{b-a}\int_a^b f(x)\,dx\)—the constant height giving same area. The Mean Value Theorem for Integrals guarantees some \(c\) where \(f(c)\) equals this average. Remember: integrals are linear (distribute over sums), but NOT over products or quotients! Always state which property you're using on AP® exams. These tools appear constantly—master them completely! 🎯✨