Unit 6.7 – The Fundamental Theorem of Calculus and Accumulation Functions

AP® Calculus AB & BC | The Bridge Between Differentiation and Integration

Why This Matters: The Fundamental Theorem of Calculus (FTC) is arguably the most important theorem in all of calculus! It reveals the profound relationship between differentiation and integration—they are inverse processes. This theorem has two parts: FTC 1 shows that differentiation "undoes" integration (accumulation functions), while FTC 2 provides a practical method to evaluate definite integrals without Riemann sums. Understanding both parts deeply is absolutely crucial for AP® Calculus success and forms the foundation for all further study in calculus!

🌟 The Big Picture: Integration and Differentiation as Inverses

THE FUNDAMENTAL CONNECTION

\[ \frac{d}{dx}\left[\int f(x)\,dx\right] = f(x) \quad \text{and} \quad \int f'(x)\,dx = f(x) + C \]

Differentiation and Integration are inverse operations!

📘 Fundamental Theorem of Calculus - Part 1

FTC Part 1 (First Fundamental Theorem)

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.

🔑 All Forms of FTC Part 1:

1. Basic Form (Upper limit is \(x\)):

\[ \frac{d}{dx}\left[\int_a^x f(t) \, dt\right] = f(x) \]

2. Chain Rule Form (Upper limit is \(u(x)\)):

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

Don't forget to multiply by \(u'(x)\)!

3. Variable Lower Limit (Lower limit is \(v(x)\)):

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

Note the negative sign!

4. Both Limits Variable:

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

📗 Fundamental Theorem of Calculus - Part 2

FTC Part 2 (Second Fundamental Theorem / Evaluation Theorem)

STATEMENT:

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

where \(F(x)\) is any antiderivative of \(f(x)\) (i.e., \(F'(x) = f(x)\))

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

Common Notation:

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

📝 Key Observations:

Any antiderivative works: The constant \(C\) cancels when subtracting
No Riemann sums needed: This makes integration practical!
Always subtract: Upper bound value minus lower bound value
Order matters: \(\int_a^b = -\int_b^a\)

📊 Accumulation Functions

ACCUMULATION FUNCTION DEFINITION

An accumulation function represents the accumulated quantity from a starting point \(a\) to a variable point \(x\):

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

Key Properties:

\(A(a) = \int_a^a f(t)\,dt = 0\) (starting value is zero)
\(A'(x) = f(x)\) (by FTC Part 1)
\(A''(x) = f'(x)\) (derivative of rate function)
\(A(x)\) represents net signed area from \(a\) to \(x\)

Analyzing Accumulation Functions:

Behavior of \(A(x)\) from Properties of \(f(x)\)
If \(f(x)\)...	Then \(A(x)\)...	Reason
\(f(x) > 0\)	is increasing	\(A'(x) = f(x) > 0\)
\(f(x) < 0\)	is decreasing	\(A'(x) = f(x) < 0\)
\(f(x) = 0\)	has critical point	\(A'(x) = 0\)
\(f\) changes from + to −	has local max	First Derivative Test
\(f\) changes from − to +	has local min	First Derivative Test
\(f\) is increasing	is concave up	\(A''(x) = f'(x) > 0\)
\(f\) is decreasing	is concave down	\(A''(x) = f'(x) < 0\)
\(f\) has max/min	has inflection point	\(A''(x) = f'(x) = 0\)

📖 Comprehensive Worked Examples

Example 1: FTC Part 1 - Basic

Problem: If \(h(x) = \int_3^x (t^2 + 2t) \, dt\), find \(h'(x)\) and \(h'(5)\).

Solution:

Apply FTC Part 1:

\[ h'(x) = x^2 + 2x \]

Simply evaluate the integrand at the upper limit \(x\)!

Evaluate at \(x = 5\):

\[ h'(5) = 5^2 + 2(5) = 25 + 10 = 35 \]

Answers: \(h'(x) = x^2 + 2x\) and \(h'(5) = 35\)

Example 2: FTC Part 1 - Chain Rule

Problem: Find \(\frac{d}{dx}\left[\int_1^{x^3} e^t \, dt\right]\).

Solution:

Identify components:

Integrand: \(f(t) = e^t\)
Upper limit: \(u(x) = x^3\)
Need Chain Rule!

Apply FTC Part 1 with Chain Rule:

\[ \frac{d}{dx}\left[\int_1^{x^3} e^t \, dt\right] = e^{x^3} \cdot \frac{d}{dx}[x^3] \]

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

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

Example 3: FTC Part 1 - Both Limits Variable

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

Solution:

Method: Split the integral

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

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

Differentiate each part:

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

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

Combine:

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

Answer: \(-\sin x \cos x(\sin x + \cos x)\) or \(\cos^2 x \sin x - \sin^2 x \cos x\)

Example 4: FTC Part 2 - Evaluation

Problem: Evaluate \(\int_1^4 (3x^2 - 2x + 1) \, dx\).

Solution:

Step 1: Find antiderivative

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

(We can verify: \(F'(x) = 3x^2 - 2x + 1\) ✓)

Step 2: Apply FTC Part 2

\[ \int_1^4 (3x^2 - 2x + 1) \, dx = \left[x^3 - x^2 + x\right]_1^4 \]

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

Step 3: Evaluate

\[ = (64 - 16 + 4) - (1 - 1 + 1) = 52 - 1 = 51 \]

Answer: 51

Example 5: Accumulation Function Analysis

Problem: Let \(g(x) = \int_0^x (t - 1)(t - 3) \, dt\).
(a) Find \(g'(x)\)
(b) Where is \(g(x)\) increasing?
(c) Where does \(g(x)\) have local extrema?
(d) Where is \(g(x)\) concave up?

Solution:

Part (a): \(g'(x)\)

\[ g'(x) = (x - 1)(x - 3) \]

Part (b): Where is \(g\) increasing?

\(g\) increasing when \(g'(x) > 0\):

\[ (x - 1)(x - 3) > 0 \]

This occurs when \(x < 1\) or \(x > 3\)

Answer: \((-\infty, 1) \cup (3, \infty)\)

Part (c): Local extrema

Critical points where \(g'(x) = 0\): \(x = 1, 3\)

At \(x = 1\): \(g'\) changes from + to − → local max
At \(x = 3\): \(g'\) changes from − to + → local min

Part (d): Concave up where \(g''(x) > 0\)

Since \(g'(x) = (x-1)(x-3) = x^2 - 4x + 3\):

\[ g''(x) = 2x - 4 \]

\[ g''(x) > 0 \text{ when } 2x - 4 > 0 \Rightarrow x > 2 \]

Answer: \((2, \infty)\)

Summary: (a) \(g'(x) = (x-1)(x-3)\) | (b) Increasing on \((-\infty,1) \cup (3,\infty)\) | (c) Local max at \(x=1\), local min at \(x=3\) | (d) Concave up on \((2,\infty)\)

💡 Essential Tips & Strategies

✅ FTC Part 1 Mastery Tips:

Check the limit: Variable in upper limit, lower limit, or both?
Upper limit formula: Evaluate integrand at upper limit, multiply by derivative of upper limit
Lower limit formula: Add negative sign, evaluate at lower limit, multiply by derivative
Both limits: Do upper minus lower (each with their derivatives)
Chain Rule is crucial: Don't forget to multiply by \(u'(x)\)!
Dummy variable: The \(t\) in \(\int f(t)\,dt\) doesn't matter—it's a placeholder

🔥 FTC Part 2 Mastery Tips:

Find ANY antiderivative: Don't add +C (it cancels anyway)
Always subtract: \(F(\text{upper}) - F(\text{lower})\)
Verify your antiderivative: Take derivative to check
Watch the signs: Common error source
Simplify before substituting: Easier calculations

🎯 Decision Tree:

If you see: \(\frac{d}{dx}\left[\int \ldots dt\right]\)

→ Use FTC Part 1

Don't integrate! Just evaluate.
Check limits, apply appropriate formula

If you see: \(\int_a^b f(x)\,dx\) (definite integral)

→ Use FTC Part 2

Find antiderivative
Evaluate at bounds and subtract

❌ Common Mistakes to Avoid

Mistake 1: Forgetting Chain Rule in FTC Part 1 when limit is not just \(x\)
Mistake 2: Missing the negative sign with variable lower limit
Mistake 3: Adding +C when using FTC Part 2 (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 which theorem to use
Mistake 7: Not simplifying \(g'(x)\) before analyzing \(g(x)\)
Mistake 8: Thinking \(f = 0\) means \(A\) has extremum without checking sign change
Mistake 9: Arithmetic errors when evaluating at bounds
Mistake 10: Not verifying antiderivative is correct

📝 Practice Problems

Set A: FTC Part 1

Find \(\frac{d}{dx}\left[\int_2^x \sin(t^2) \, dt\right]\)
Find \(\frac{d}{dx}\left[\int_0^{x^2} \sqrt{1+t^3} \, dt\right]\)
Find \(\frac{d}{dx}\left[\int_{x}^5 e^{t^2} \, dt\right]\)

Answers:

\(\sin(x^2)\)
\(\sqrt{1+(x^2)^3} \cdot 2x = 2x\sqrt{1+x^6}\)
\(-e^{x^2}\)

Set B: FTC Part 2

\(\int_0^2 (x^3 - 4x) \, dx\)
\(\int_1^e \frac{1}{x} \, dx\)

Answers:

\(\left[\frac{x^4}{4} - 2x^2\right]_0^2 = (4-8)-(0) = -4\)
\([\ln|x|]_1^e = \ln e - \ln 1 = 1 - 0 = 1\)

Set C: Accumulation Functions

If \(F(x) = \int_1^x t^2\,dt\), find where \(F\) is increasing

Answer:

\(F'(x) = x^2 > 0\) for all \(x \neq 0\), so increasing on \((-\infty, 0) \cup (0, \infty)\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

State which FTC: "By FTC Part 1..." or "By FTC Part 2..."
Show Chain Rule: Write \(f(u(x)) \cdot u'(x)\) explicitly
Show antiderivative: Write \(F(x)\) before evaluating
Use bracket notation: \([F(x)]_a^b\) is clear
Show subtraction: \(F(b) - F(a)\) step
Justify extrema: Show sign change of derivative
Units when applicable: Context problems need units

⚡ Ultimate Quick Reference

THE FUNDAMENTAL THEOREM - COMPLETE SUMMARY

Theorem	Formula	Use When
FTC Part 1 (Basic)	\(\frac{d}{dx}\left[\int_a^x f(t)\,dt\right] = f(x)\)	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 1 (Lower)	\(\frac{d}{dx}\left[\int_{v(x)}^a f(t)\,dt\right] = -f(v(x)) \cdot v'(x)\)	Lower limit variable
FTC Part 2	\(\int_a^b f(x)\,dx = F(b) - F(a)\)	Evaluating integral

The FTC Changes Everything! The Fundamental Theorem of Calculus is the most important result in calculus, revealing that differentiation and integration are inverse operations. FTC Part 1 says \(\frac{d}{dx}\left[\int_a^x f(t)\,dt\right] = f(x)\)—differentiating an accumulation function gives back the rate function. When the 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 and subtract 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 \(f\) has extrema, \(A\) has inflection points. Always state which part of FTC you're using, show Chain Rule when needed, and verify antiderivatives. This theorem transforms integration from theoretical Riemann sums into practical computation. Master it completely—it's tested extensively on every AP® Calculus exam! 🎯✨