Mathematics Advanced • Year 12 • Module 6 • Lesson 5

The Fundamental Theorem of Calculus

Practise HSC-style writing on both parts of the FTC and the connection between differentiation and integration.

Master · Past-Paper Style

1. Short-answer questions

1.1 Evaluate ∫₀²(x³ + 3x² − 2x + 1) dx using FTC Part 2. Show all working. 3 marks Band 3-4

1.2 Find d/dx (∫₁^x² (t³ + 1) dt). Show your application of the chain rule. 3 marks Band 4

1.3 Let g(x) = ∫₀^x (4t − t²) dt for x ≥ 0.
(a) Find g'(x) using FTC Part 1.
(b) Find the value(s) of x at which g(x) is stationary.
(c) Use FTC Part 2 to find g(x) explicitly, and confirm your stationary point(s) using your formula. 4 marks Band 4-5

Stuck on 1.3(b)? Stationary points occur where g'(x) = 0; by FTC Part 1 this is where the integrand is zero.

2. Extended response

2.1 The Fundamental Theorem of Calculus is often called the most important theorem in mathematics.
(a) State both parts of the FTC precisely.
(b) Use FTC Part 2 to evaluate ∫₁^e(1/x) dx and ∫₀⁴(3x² + 2) dx, showing all working.
(c) For the integral function g(x) = ∫₂^x² e^t dt, use FTC Part 1 with the chain rule to find g'(x); state the value of g'(1).
(d) In two sentences, explain why the FTC is considered to unify differentiation and integration into a single theory — referring explicitly to the inverse-operation relationship and the area-problem context. 8 marks Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — Part 1: d/dx (∫_a^x f(t) dt) = f(x) for f continuous on [a, b].

• 1 mark — Part 2: ∫_a^b f(x) dx = F(b) − F(a), where F is any antiderivative of f.

Part (b) — 2 marks

• 1 mark — ∫₁^e(1/x) dx = [ln|x|]₁^e = ln e − ln 1 = 1.

• 1 mark — ∫₀⁴(3x² + 2) dx = [x³ + 2x]₀⁴ = 64 + 8 = 72.

Part (c) — 2 marks

• 1 mark — g'(x) = e^x² · 2x = 2x e^x².

• 1 mark — g'(1) = 2 e¹ = 2e.

Part (d) — 2 marks

• 1 mark — articulates the inverse-operation relationship (differentiating an integral function returns the integrand; integrating a derivative returns the function up to + C).

• 1 mark — connects this to the area-problem: before FTC, areas required summing infinitely many rectangles; the FTC reduces this to evaluating an antiderivative at two points.

Your response:

Stuck on (d)? Two sentences: (1) inverse-operation pairing; (2) "area-problem" — Riemann sums → antiderivative shortcut.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — ∫₀²(x³ + 3x² − 2x + 1) dx (3 marks)

Sample response. F(x) = x⁴/4 + x³ − x² + x. [F(x)]₀² = (16/4 + 8 − 4 + 2) − 0 = 4 + 8 − 4 + 2 = 10.

Marking notes. 1 mark — correct antiderivative. 1 mark — correct substitution at upper and lower limits. 1 mark — final value 10 with no + C in the definite-integral answer.

1.2 — d/dx (∫₁^x²(t³ + 1) dt) (3 marks)

Sample response. By FTC Part 1 with u(x) = x², u'(x) = 2x: d/dx (∫₁^x²(t³ + 1) dt) = ((x²)³ + 1) · 2x = (x⁶ + 1) · 2x = 2x⁷ + 2x.

Marking notes. 1 mark — names FTC Part 1 / identifies the chain-rule structure. 1 mark — correctly substitutes u = x² into the integrand. 1 mark — multiplies by u'(x) = 2x and simplifies. Forgetting the chain factor 2x is the most common error and caps the response at 1/3.

1.3 — g(x) = ∫₀^x(4t − t²) dt (4 marks)

Sample response.

(a) By FTC Part 1, g'(x) = 4x − x².
(b) g'(x) = 0 ⇒ x(4 − x) = 0 ⇒ x = 0 or x = 4. Stationary points at x = 0 and x = 4.
(c) g(x) = ∫₀^x(4t − t²) dt = [2t² − t³/3]₀^x = 2x² − x³/3. Differentiating directly: g'(x) = 4x − x², which equals 0 at x = 0 and x = 4 ✓ — confirms (a) and (b).

Marking notes. (a) 1 mark — g'(x) by FTC Part 1. (b) 1 mark — solves g' = 0 and states both roots. (c) 2 marks — 1 for the explicit g(x) via Part 2, 1 for the verification by direct differentiation.

2.1 — Extended response (8 marks): sample Band-6 response with annotations

Sample Band-6 response.

(a) Let f be a function continuous on [a, b], and let F be any antiderivative of f.
FTC Part 1. The integral function g(x) = ∫_a^x f(t) dt is differentiable on (a, b), and g'(x) = f(x). [1 mark.]
FTC Part 2. ∫_a^b f(x) dx = F(b) − F(a). [1 mark.]

(b)

∫₁^e(1/x) dx = [ln|x|]₁^e = ln e − ln 1 = 1 − 0 = 1. [1 mark.]

∫₀⁴(3x² + 2) dx = [x³ + 2x]₀⁴ = (64 + 8) − 0 = 72. [1 mark.]

(c) By FTC Part 1 with u(x) = x², u'(x) = 2x:

g'(x) = e^u(x) · u'(x) = e^x² · 2x = 2x e^x². [1 mark — derivative with chain factor.]

g'(1) = 2 · 1 · e¹ = 2e. [1 mark.]

(d) The FTC unifies differentiation and integration by showing they are inverse operations: differentiating an integral function (Part 1) returns the integrand, and integrating the derivative of a function (Part 2) recovers that function up to a constant. [1 mark.] This is also what makes the FTC powerful in practice: a centuries-old open problem (finding areas under curves, which previously required summing infinitely many rectangles via Riemann sums) is reduced by Part 2 to the routine task of finding an antiderivative and evaluating it at two points. [1 mark.]

Total: 8/8.

Band descriptors for marker.

Band 3: States either Part 1 or Part 2 (not both); completes one of the two integrals in (b); little or no progress on (c)(d). ≈ 2-3 marks.

Band 4: Both parts in (a); both integrals in (b); attempts (c) but misses the chain factor (writes e^x² only); (d) is vague. ≈ 4-5 marks.

Band 5: Full marks on (a)(b)(c); (d) discusses inverse operations but does not mention the area-problem / Riemann-sum context. ≈ 6-7 marks.

Band 6: Full working in all four parts, precise definitions in (a), explicit chain rule in (c), and (d) covers both the inverse-operation pairing and the area-problem reduction. 8/8.