Mathematics Advanced • Year 12 • Module 6 • Lesson 5

The Fundamental Theorem of Calculus

Apply both parts of the FTC to accumulation, kinematics and economic problems, including chain-rule differentiation of integral functions.

Apply · Problem Set

Problem 1 — Capital accumulation (economics)

An economy's investment rate (in $billion/year, t years after 2026) is

I(s) = 50 + 4 s, s ≥ 0.

The capital stock at time t is K(t) = ∫₀^t I(s) ds (with K(0) = 0 baseline).

Set up: What are we solving for?

(i) Use FTC Part 2 to evaluate K(t) as an explicit function of t. 2 marks

(ii) Use FTC Part 1 to write down dK/dt and interpret it in one sentence. 2 marks

(iii) Find the capital stock at t = 10 (i.e. K(10)). 1 mark

Stuck? Revisit lesson § Real-World Anchor — Accumulation Functions in Economics.

Problem 2 — Speed → distance via FTC (kinematics)

A drone's speed (always positive) along a straight rail is

v(t) = 2 + 3 √t m/s, t ≥ 0.

Let s(t) = ∫₀^t v(τ) dτ be the distance travelled.

Set up: What are we solving for?

(i) Use FTC Part 1 to confirm ds/dt = v(t). 1 mark

(ii) Use FTC Part 2 to find s(t) explicitly. 2 marks

(iii) Calculate s(4) and verify your formula by also computing ∫₀⁴ v(t) dt directly. 3 marks

Problem 3 — Chain-rule integral functions

Define g(x) = ∫₁^x³ (t² + t) dt for x > 0.

Set up: What are we solving for?

(i) Use FTC Part 1 with the chain rule to find g'(x). 2 marks

(ii) Find g'(2). 1 mark

(iii) Use FTC Part 2 to find g(x) as an explicit polynomial in x, then differentiate it directly and verify the result agrees with (i). 3 marks

Stuck on (iii)? Antiderivative is F(t) = t³/3 + t²/2; g(x) = F(x³) − F(1).

Problem 4 — Area under a curve via FTC

For a long time mathematicians struggled to find areas; the FTC turned this into a routine antiderivative problem.

Set up: What are we solving for?

(i) Find the area between y = x² and the x-axis from x = 0 to x = 3 using FTC Part 2. 2 marks

(ii) Find the area between y = e^x and the x-axis from x = 0 to x = 2 (exact form). 2 marks

(iii) Comment in one sentence on how the FTC reduces the area problem (in principle requiring Riemann sums of infinitely many rectangles) to evaluating an antiderivative at two points. 2 marks

Problem 5 — Using Part 1 to check antiderivatives

FTC Part 1 gives a quick check on whether a proposed antiderivative is correct: differentiate it and compare with the integrand.

Set up: What are we solving for?

(i) Verify that F(x) = (1/2) e^2x is an antiderivative of f(x) = e^2x, hence evaluate ∫₀¹ e^2x dx in exact form. 3 marks

(ii) A student writes "G(x) = x ln x is an antiderivative of ln x." Differentiate G and identify the missing term, then state the correct antiderivative of ln x. 3 marks

Stuck on (ii)? Apply the product rule: d/dx(x ln x) = ln x + 1.

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Answers — Do not peek before attempting

Problem 1 — Capital accumulation

Set up. Apply FTC Part 2 to express K(t) in closed form, then FTC Part 1 to recover the rate, and finally evaluate K(10).

(i) K(t) = ∫₀^t(50 + 4s) ds = [50s + 2s²]₀^t = 50 t + 2 t².

(ii) dK/dt = 50 + 4t = I(t) by FTC Part 1 — the rate of change of capital equals the investment rate at time t.

(iii) K(10) = 500 + 200 = $700 billion.

Problem 2 — Speed → distance

Set up. FTC Part 1 confirms s'(t) = v(t); FTC Part 2 produces the closed-form distance, then both routes are verified to agree at t = 4.

(i) By FTC Part 1, ds/dt = v(t) = 2 + 3√t ✓ (the integrand re-emerges as the derivative).

(ii) s(t) = ∫₀^t(2 + 3 τ^1/2) dτ = [2τ + 2 τ^3/2]₀^t = 2t + 2 t^3/2.

(iii) s(4) = 8 + 2 · 8 = 24 m. Directly: ∫₀⁴(2 + 3√t) dt = [2t + 2 t^3/2]₀⁴ = (8 + 16) − 0 = 24 ✓ — agreement confirms the two routes are equivalent.

Problem 3 — Chain-rule integral function

Set up. Use FTC Part 1 with u(x) = x³ to get g'(x), evaluate at x = 2, then independently compute g(x) explicitly and differentiate to cross-check.

(i) g'(x) = ((x³)² + x³) · 3x² = (x⁶ + x³) · 3x² = 3x⁸ + 3x⁵.

(ii) g'(2) = 3 · 256 + 3 · 32 = 768 + 96 = 864.

(iii) F(t) = t³/3 + t²/2. g(x) = F(x³) − F(1) = (x³)³/3 + (x³)²/2 − (1/3 + 1/2) = x⁹/3 + x⁶/2 − 5/6. Differentiating: dg/dx = 3x⁸ + 3x⁵ ✓ (matches part (i)).

Problem 4 — Area under a curve

Set up. Apply FTC Part 2 to each integral.

(i) ∫₀³ x² dx = [x³/3]₀³ = 27/3 − 0 = 9 square units.

(ii) ∫₀² e^x dx = [e^x]₀² = e² − 1.

(iii) Before the FTC, finding the area under a curve required summing infinitely many rectangles (Riemann sums). The FTC replaces that infinite process with a single antiderivative and two substitutions — a method-of-the-millennium shortcut that turned a centuries-old open problem into a routine calculation.

Problem 5 — Using Part 1 to check antiderivatives

Set up. Differentiate the proposed antiderivative; if it equals f(x), then F is correct and the definite integral is just F(b) − F(a). Otherwise the discrepancy points to the correction needed.

(i) d/dx ((1/2) e^2x) = (1/2) · 2 · e^2x = e^2x ✓. Hence ∫₀¹ e^2x dx = [(1/2) e^2x]₀¹ = (e² − 1)/2 = (e² − 1)/2.

(ii) d/dx (x ln x) = ln x + x · (1/x) = ln x + 1 (not ln x — there's an extra + 1). To remove the extra 1, subtract x: the correct antiderivative of ln x is x ln x − x + C (since d/dx(x ln x − x) = ln x + 1 − 1 = ln x ✓).