Unit 6.4 – The Fundamental Theorem of Calculus and Accumulation Functions

AP® Calculus AB & BC | The Most Important Theorem in Calculus!

Why This Matters: The Fundamental Theorem of Calculus (FTC) is the crown jewel of calculus! It establishes the profound connection between differentiation and integration—they are inverse operations. This single theorem revolutionized mathematics and science by showing that finding areas (integration) can be done by reversing derivatives (antiderivatives). It has two parts: FTC Part 1 defines accumulation functions and shows differentiation "undoes" integration, while FTC Part 2 gives us a practical way to evaluate definite integrals without limits of Riemann sums. Mastering this theorem is absolutely essential for AP® Calculus success!

🎯 The Fundamental Theorem of Calculus - Part 1

FTC Part 1 (The Accumulation Theorem)

THE STATEMENT:

\[ \text{If } g(x) = \int_a^x f(t) \, dt, \text{ then } g'(x) = f(x) \]

In words: The derivative of an accumulation function equals the integrand evaluated at the upper limit.

More Generally (Chain Rule Version):

\[ \frac{d}{dx}\left[\int_a^{u(x)} f(t) \, dt\right] = f(u(x)) \cdot u'(x) \]

If upper limit is a function of \(x\), use Chain Rule!

With Variable Lower Limit:

\[ \frac{d}{dx}\left[\int_{u(x)}^a f(t) \, dt\right] = -f(u(x)) \cdot u'(x) \]

Variable lower limit introduces negative sign

Both Limits Variable:

\[ \frac{d}{dx}\left[\int_{u(x)}^{v(x)} f(t) \, dt\right] = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x) \]

Split into two integrals, apply FTC Part 1 to each

🔑 Key Insights from FTC Part 1:

Differentiation undoes integration: \(\frac{d}{dx}\int f(x)\,dx = f(x)\)
Accumulation functions are antiderivatives
The lower limit is a constant: It doesn't affect the derivative
Upper limit is the variable: Evaluate integrand there
Don't forget Chain Rule: If upper limit is \(u(x)\), multiply by \(u'(x)\)

⚡ The Fundamental Theorem of Calculus - Part 2

FTC Part 2 (The Evaluation Theorem)

THE STATEMENT:

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

where \(F(x)\) is any antiderivative of \(f(x)\)

In words: To evaluate a definite integral, find an antiderivative, then subtract its values at the bounds.

Common Notation:

\[ \int_a^b f(x) \, dx = F(x)\Big|_a^b = \left[F(x)\right]_a^b = F(b) - F(a) \]

All three notations mean the same thing!

🔑 Key Insights from FTC Part 2:

No need for Riemann sums anymore! Just find antiderivative
Any antiderivative works: The constant \(C\) cancels out
Always subtract: Upper limit minus lower limit
Order matters: \(\int_a^b = -\int_b^a\)
Makes integration practical: This is how we actually compute integrals

📈 Accumulation Functions

DEFINITION

An accumulation function is defined by an integral with a variable upper limit:

\[ A(x) = \int_a^x f(t) \, dt \]

This represents the accumulated area under \(f(t)\) from \(a\) to \(x\)

Key Properties of Accumulation Functions:

\(A(a) = \int_a^a f(t)\,dt = 0\) (zero width = zero area)
\(A'(x) = f(x)\) (by FTC Part 1)
When \(f(x) > 0\): \(A(x)\) is increasing
When \(f(x) < 0\): \(A(x)\) is decreasing
When \(f(x)\) changes sign: \(A(x)\) has extrema
\(A''(x) = f'(x)\) (take derivative again)

📖 Comprehensive Worked Examples

Example 1: FTC Part 2 - Basic Evaluation

Problem: Evaluate \(\int_1^4 (2x + 3) \, dx\)

Solution:

Step 1: Find antiderivative

Antiderivative of \(2x + 3\) is \(x^2 + 3x\) (plus constant, but it cancels)

\[ F(x) = x^2 + 3x \]

Step 2: Apply FTC Part 2

\[ \int_1^4 (2x + 3) \, dx = \left[x^2 + 3x\right]_1^4 \]

\[ = F(4) - F(1) \]

Step 3: Evaluate at bounds

\[ = (4^2 + 3 \cdot 4) - (1^2 + 3 \cdot 1) \]

\[ = (16 + 12) - (1 + 3) = 28 - 4 = 24 \]

Answer: \(\int_1^4 (2x + 3) \, dx = 24\)

Example 2: FTC Part 1 - Simple Accumulation Function

Problem: If \(g(x) = \int_2^x t^2 \, dt\), find \(g'(x)\) and \(g'(3)\).

Solution:

Apply FTC Part 1:

The derivative of \(\int_a^x f(t)\,dt\) is \(f(x)\)

\[ g'(x) = x^2 \]

Evaluate at \(x = 3\):

\[ g'(3) = 3^2 = 9 \]

Answer: \(g'(x) = x^2\) and \(g'(3) = 9\)

Example 3: FTC Part 1 with Chain Rule

Problem: Find \(\frac{d}{dx}\left[\int_0^{x^2} \sin(t) \, dt\right]\)

Solution:

Identify the setup:

Integrand: \(f(t) = \sin(t)\)
Upper limit: \(u(x) = x^2\)
Need Chain Rule!

Apply FTC Part 1 with Chain Rule:

\[ \frac{d}{dx}\left[\int_0^{x^2} \sin(t) \, dt\right] = \sin(x^2) \cdot \frac{d}{dx}[x^2] \]

\[ = \sin(x^2) \cdot 2x = 2x\sin(x^2) \]

Answer: \(2x\sin(x^2)\)

Example 4: Variable Lower Limit

Problem: Find \(\frac{d}{dx}\left[\int_{x^3}^5 e^t \, dt\right]\)

Solution:

Method 1: Use property \(\int_a^b = -\int_b^a\)

\[ \int_{x^3}^5 e^t \, dt = -\int_5^{x^3} e^t \, dt \]

Now upper limit is \(x^3\):

\[ \frac{d}{dx}\left[-\int_5^{x^3} e^t \, dt\right] = -e^{x^3} \cdot 3x^2 = -3x^2e^{x^3} \]

Method 2: Direct formula

Variable lower limit → negative sign:

\[ \frac{d}{dx}\left[\int_{u(x)}^a f(t) \, dt\right] = -f(u(x)) \cdot u'(x) \]

\[ = -e^{x^3} \cdot 3x^2 = -3x^2e^{x^3} \]

Answer: \(-3x^2e^{x^3}\)

Example 5: Both Limits Variable

Problem: Find \(\frac{d}{dx}\left[\int_{x^2}^{2x} \cos(t) \, dt\right]\)

Solution:

Split into two integrals:

\[ \int_{x^2}^{2x} \cos(t) \, dt = \int_{x^2}^0 \cos(t) \, dt + \int_0^{2x} \cos(t) \, dt \]

\[ = -\int_0^{x^2} \cos(t) \, dt + \int_0^{2x} \cos(t) \, dt \]

Differentiate each part:

\[ \frac{d}{dx}\left[-\int_0^{x^2} \cos(t) \, dt\right] = -\cos(x^2) \cdot 2x \]

\[ \frac{d}{dx}\left[\int_0^{2x} \cos(t) \, dt\right] = \cos(2x) \cdot 2 \]

Combine:

\[ -2x\cos(x^2) + 2\cos(2x) = 2\cos(2x) - 2x\cos(x^2) \]

Answer: \(2\cos(2x) - 2x\cos(x^2)\)

Example 6: Analyzing Accumulation Function

Problem: Let \(F(x) = \int_0^x (t^2 - 4) \, dt\).
(a) Find \(F'(x)\)
(b) For what values of \(x\) is \(F(x)\) increasing?
(c) At what values of \(x\) does \(F(x)\) have extrema?

Solution:

Part (a): Apply FTC Part 1

\[ F'(x) = x^2 - 4 \]

Part (b): \(F\) increasing when \(F'(x) > 0\)

\[ x^2 - 4 > 0 \]

\[ (x-2)(x+2) > 0 \]

Solution: \(x < -2\) or \(x > 2\)

Part (c): Extrema when \(F'(x) = 0\)

\[ x^2 - 4 = 0 \quad \Rightarrow \quad x = \pm 2 \]

Since \(F'(x)\) changes from + to − at \(x = -2\): local max

Since \(F'(x)\) changes from − to + at \(x = 2\): local min

Answers: (a) \(F'(x) = x^2 - 4\) | (b) \((-\infty, -2) \cup (2, \infty)\) | (c) Local max at \(x = -2\), local min at \(x = 2\)

🔧 Essential Properties and Rules

Key Properties Summary:

Definite Integral Properties
Property	Formula
Zero width	\(\int_a^a f(x)\,dx = 0\)
Reverse limits	\(\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx\)
Constant multiple	\(\int_a^b cf(x)\,dx = c\int_a^b f(x)\,dx\)
Sum/Difference	\(\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx\)
Additivity	\(\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx\)

💡 Essential Tips & Strategies

✅ FTC Part 1 Tips:

Check which limit is variable: Upper, lower, or both?
Upper limit variable: Evaluate integrand at upper limit
Lower limit variable: Add negative sign
Don't forget Chain Rule: If limit is \(u(x)\), multiply by \(u'(x)\)
Lower limit doesn't matter: It's just a constant
Can't integrate the function? No problem for FTC Part 1!

🔥 FTC Part 2 Tips:

Find ANY antiderivative: Don't add +C (it cancels)
Always subtract: Upper − Lower
Check your antiderivative: Differentiate it to verify
Be careful with signs: Especially with negative functions
Units matter: In applied problems, state units

🎯 Decision Tree for FTC Problems:

If asked to DIFFERENTIATE an integral: Use FTC Part 1

Check if limits have variables
Apply appropriate formula
Don't forget Chain Rule!

If asked to EVALUATE an integral: Use FTC Part 2

Find antiderivative
Evaluate at upper limit
Subtract value at lower limit

❌ Common Mistakes to Avoid

Mistake 1: Forgetting Chain Rule when upper limit is a function
Mistake 2: Not adding negative sign for variable lower limit
Mistake 3: Adding +C when using FTC Part 2 (it's not needed!)
Mistake 4: Subtracting in wrong order: must be \(F(b) - F(a)\)
Mistake 5: Trying to integrate when using FTC Part 1 (just evaluate!)
Mistake 6: Confusing variable of integration \(t\) with limit variable \(x\)
Mistake 7: Not verifying antiderivative by differentiating
Mistake 8: Forgetting to evaluate antiderivative at BOTH limits
Mistake 9: Sign errors with negative functions
Mistake 10: Not simplifying expressions before taking derivative

📝 Practice Problems

Set A: FTC Part 2 (Evaluation)

\(\int_0^3 (x^2 + 2x) \, dx\)
\(\int_1^4 \frac{1}{x} \, dx\)
\(\int_0^{\pi} \sin(x) \, dx\)

Answers:

\(\left[\frac{x^3}{3} + x^2\right]_0^3 = (9 + 9) - 0 = 18\)
\([\ln|x|]_1^4 = \ln 4 - \ln 1 = \ln 4\)
\([-\cos x]_0^{\pi} = -(-1) - (-1) = 2\)

Set B: FTC Part 1 (Derivatives)

\(\frac{d}{dx}\left[\int_3^x t^3 \, dt\right]\)
\(\frac{d}{dx}\left[\int_1^{x^2} e^t \, dt\right]\)
\(\frac{d}{dx}\left[\int_{\sqrt{x}}^4 \cos(t) \, dt\right]\)

Answers:

\(x^3\)
\(e^{x^2} \cdot 2x\)
\(-\cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\)

Set C: Accumulation Functions

If \(g(x) = \int_0^x (t-2)(t+1) \, dt\), find where \(g(x)\) has local extrema

Answer:

\(g'(x) = (x-2)(x+1) = 0\) at \(x = -1, 2\); local max at \(x = -1\), local min at \(x = 2\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Clear identification: State which part of FTC you're using
Show antiderivative: Write \(F(x)\) explicitly for Part 2
Chain Rule shown: For Part 1 with composite limits
Evaluation notation: Use brackets with limits: \([F(x)]_a^b\)
Subtraction shown: Write \(F(b) - F(a)\) explicitly
Units included: For applied/context problems
Justification: Explain why extrema occur

💯 Maximizing Your Score:

Show all work: Even "easy" steps—partial credit available
Use correct notation: \(\int_a^b\), not \(\int^b_a\)
Don't skip substitution: Show evaluation at each limit
Verify reasonableness: Does your answer make sense?

⚡ Ultimate Quick Reference

THE FUNDAMENTAL THEOREM - BOTH PARTS

Part	Formula	Use When
FTC Part 1	\(\frac{d}{dx}\left[\int_a^x f(t)\,dt\right] = f(x)\)	Differentiating integral
FTC Part 1 + Chain	\(\frac{d}{dx}\left[\int_a^{u(x)} f(t)\,dt\right] = f(u(x)) \cdot u'(x)\)	Upper limit is \(u(x)\)
FTC Part 2	\(\int_a^b f(x)\,dx = F(b) - F(a)\)	Evaluating integral

Master the FTC! The Fundamental Theorem of Calculus is THE most important result in calculus. FTC Part 1 says \(\frac{d}{dx}\left[\int_a^x f(t)\,dt\right] = f(x)\)—differentiation undoes integration! When the upper limit is \(u(x)\), use Chain Rule: multiply by \(u'(x)\). For variable lower limits, add a negative sign. FTC Part 2 says \(\int_a^b f(x)\,dx = F(b) - F(a)\)—to evaluate a definite integral, find any antiderivative \(F(x)\), then subtract its values at the bounds. No need for +C since it cancels! Accumulation functions \(A(x) = \int_a^x f(t)\,dt\) have \(A'(x) = f(x)\), so where \(f > 0\), \(A\) increases; where \(f < 0\), \(A\) decreases. These theorems transform integration from tedious Riemann sum limits into straightforward antiderivative evaluation. Always check: differentiating? Use Part 1. Evaluating? Use Part 2. Don't forget Chain Rule for composite limits! This theorem is tested extensively on AP® exams—master it completely! 🎯✨