Mathematics Advanced • Year 12 • Module 6 • Lesson 5

The Fundamental Theorem of Calculus

Build fluency in both parts of the FTC: evaluating definite integrals via F(b) − F(a), and differentiating integral functions with the chain rule.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the two parts of the FTC:

Part 1: d/dx ( ∫_a^x f(t) dt ) = ____________

Part 2: ∫_a^b f(x) dx = ____________ − ____________, where F is any antiderivative of f.

Q1.2 Complete the chain-rule extension: d/dx ( ∫_a^u(x) f(t) dt ) = ____________ · ____________.

Q1.3 In one sentence, what does the FTC tell us about the relationship between differentiation and integration? ____________________________________________________________

Stuck? Revisit lesson § Formula Reference.

2. Worked example — d/dx ( ∫₁^x² ln t dt )

Follow every line. Each step has a reason on the right.

Problem. Find d/dx ( ∫₁^x² ln t dt ) for x > 0.

Step 1 — Identify the integrand and the upper limit.

f(t) = ln t, u(x) = x²

Reason: the lesson form is ∫_a^u(x) f(t) dt with u(x) on the upper limit.

Step 2 — Apply FTC Part 1 with chain rule: f(u(x)) · u'(x).

f(u(x)) = ln(x²)

u'(x) = 2x

Reason: differentiate the upper limit only; the lower limit a is constant so contributes 0.

Step 3 — Multiply.

d/dx (∫_1^{x²} ln t dt) = ln(x²) · 2x

= 2x · 2 ln x = 4x ln x (using ln(x²) = 2 ln x for x > 0)

Answer. 4x ln x (for x > 0).

3. Faded example — fill in the missing steps

Find d/dx ( ∫₀^3x e^t² dt ). Fill in each blank line. 4 marks

Step 1 — Identify f and u:

f(t) = ____________ u(x) = ____________

Step 2 — Compute f(u(x)) and u'(x):

f(u(x)) = ____________ u'(x) = ____________

Step 3 — Multiply:

d/dx ( ∫₀^3x e^t² dt ) = ____________ · ____________ = ____________

Stuck? Substitute u into f, then multiply by u'(x); no integration of e^t² is required.

4. Graduated practice — evaluate or differentiate

Use FTC Part 1 (differentiating an integral) or Part 2 (evaluating a definite integral) as appropriate.

Foundation — straight FTC Part 1 or Part 2 (4 questions)

Q	Expression	Answer
4.1 1	∫₀⁴ x³ dx
4.2 1	d/dx (∫₂^x √t dt)
4.3 1	d/dx (∫₀^x e^t dt)
4.4 1	∫₁^e (1/x) dx

Standard — typical HSC difficulty (6 questions)

Show working on the line below each part.

4.5 ∫₀²(x³ + 3x² − 2x + 1) dx 2 marks

4.6 d/dx (∫₀^x³ e^t dt) 2 marks

4.7 d/dx (∫₁^x² (t³ + 1) dt) 2 marks

4.8 ∫₀¹(e^x + x²) dx (exact) 2 marks

4.9 Verify FTC Part 1 directly by computing d/dx (∫₀^x 2t dt) and showing it equals 2x. 2 marks

4.10 If K(t) = ∫₀^t(3s + 2) ds models capital accumulation, find dK/dt and interpret it. 2 marks

Extension — combine both parts (2 questions)

4.11 Let g(x) = ∫₀^x² sin t dt. (a) State g'(x) using Part 1 with chain rule. (b) Evaluate g(0) using Part 2. 3 marks

4.12 A student claims "d/dx (∫₀^x² f(t) dt) = f(x)" for any continuous f. Identify the missing factor and explain in one sentence why it must be there. 3 marks

Stuck on 4.12? The upper limit is u(x) = x², not x; chain rule supplies u'(x) = 2x.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 (∫₀⁴ x³ dx) I evaluated [x⁴/4]₀⁴ = 64 − 0 = 64 (and did not include + C).

For 4.2 (d/dx ∫₂^x √t dt) I wrote √x directly (the lower constant doesn't matter; chain factor is 1 since upper limit is just x).

For 4.3 (d/dx ∫₀^x e^t dt) I wrote e^x directly; no chain factor needed.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Two parts

Part 1: d/dx ( ∫_a^x f(t) dt ) = f(x). Part 2: ∫_a^b f(x) dx = F(b) − F(a).

Q1.2 — Chain-rule version

d/dx ( ∫_a^u(x) f(t) dt ) = f(u(x)) · u'(x).

Q1.3 — Inverse relationship

Differentiation and integration are inverse operations — integrating a function and then differentiating recovers the original function (up to + C for indefinite integrals).

Q3 — Faded example d/dx(∫₀^3x e^t² dt)

Step 1: f(t) = e^t²; u(x) = 3x.
Step 2: f(u(x)) = e^(3x)² = e^9x²; u'(x) = 3.
Step 3: e^9x² · 3 = 3 e^9x².

Q4.1 — ∫₀⁴ x³ dx

[x⁴/4]₀⁴ = 64 − 0 = 64.

Q4.2 — d/dx (∫₂^x √t dt)

√x by FTC Part 1 (upper limit is x; u'(x) = 1).

Q4.3 — d/dx (∫₀^x e^t dt)

e^x.

Q4.4 — ∫₁^e (1/x) dx

[ln|x|]₁^e = ln e − ln 1 = 1.

Q4.5 — ∫₀²(x³ + 3x² − 2x + 1) dx

F(x) = x⁴/4 + x³ − x² + x. F(2) = 16/4 + 8 − 4 + 2 = 4 + 8 − 4 + 2 = 10. F(0) = 0. Answer: 10.

Q4.6 — d/dx (∫₀^x³ e^t dt)

e^u · u' = e^x³ · 3x² = 3x² e^x³.

Q4.7 — d/dx (∫₁^x² (t³ + 1) dt)

((x²)³ + 1) · 2x = (x⁶ + 1) · 2x = 2x⁷ + 2x.

Q4.8 — ∫₀¹(e^x + x²) dx

[e^x + x³/3]₀¹ = (e + 1/3) − (1 + 0) = e − 2/3.

Q4.9 — Verify FTC Part 1 for ∫₀^x 2t dt

∫₀^x 2t dt = [t²]₀^x = x². d/dx (x²) = 2x ✓ — matches the integrand at the upper limit, as Part 1 predicts.

Q4.10 — Capital accumulation

By FTC Part 1, dK/dt = 3t + 2. Interpretation: the instantaneous rate at which capital is accumulating at time t equals the original investment rate function evaluated at t.

Q4.11 — g(x) = ∫₀^x² sin t dt

(a) g'(x) = sin(x²) · 2x = 2x sin(x²). (b) g(0) = ∫₀⁰ sin t dt = 0 (same limits ⇒ integral is 0).

Q4.12 — Missing chain factor

Correct: d/dx ( ∫₀^x² f(t) dt ) = f(x²) · 2x. The missing factor is u'(x) = 2x. It must be there because the chain rule treats the integral as a composition (outer = integral function with upper limit u, inner = u(x)); differentiating the inner function gives u'(x) = 2x.

1. Quick recall

2. Worked example — d/dx ( ∫1x² ln t dt )

3. Faded example — fill in the missing steps

4. Graduated practice — evaluate or differentiate

Foundation — straight FTC Part 1 or Part 2 (4 questions)

Standard — typical HSC difficulty (6 questions)

Extension — combine both parts (2 questions)

5. Self-check the easy 3

Q1.1 — Two parts

Q1.2 — Chain-rule version

Q1.3 — Inverse relationship

Q3 — Faded example d/dx(∫03x et² dt)

Q4.1 — ∫04 x³ dx

Q4.2 — d/dx (∫2x √t dt)

Q4.3 — d/dx (∫0x et dt)

Q4.4 — ∫1e (1/x) dx

Q4.5 — ∫02(x³ + 3x² − 2x + 1) dx

Q4.6 — d/dx (∫0x³ et dt)

Q4.7 — d/dx (∫1x² (t³ + 1) dt)

Q4.8 — ∫01(ex + x²) dx

Q4.9 — Verify FTC Part 1 for ∫0x 2t dt