Mathematics Advanced • Year 12 • Module 6 • Lesson 3

Integrating Exponentials and Logarithms

Practise HSC-style writing on exponential and reciprocal antiderivatives, including a long-form analysis of a tracer-decay problem.

Master · Past-Paper Style

1. Short-answer questions

1.1 Find ∫ (3 e^2x − 5/x + 4x) dx. Show all working. 3 marks Band 3-4

1.2 Find ∫ (e^−x + x⁻¹ + x^1/2) dx. Show all working. 3 marks Band 4

1.3 Show that d/dx (e^kx/k) = e^kx for any non-zero constant k. Hence write down ∫ e^−3x dx and ∫ 4 e^0.5x dx. 4 marks Band 4

Stuck on 1.3? Apply the chain rule to e^kx: d/dx(e^kx) = k e^kx.

2. Extended response

2.1 A radioactive tracer in a patient's bloodstream decays at rate dA/dt = −2 e^{−0.5 t} mg/h, with A(0) = 4 mg.
(a) Find A(t).
(b) Calculate the amount remaining after 4 hours, to 2 d.p.
(c) Describe the long-term behaviour of A(t) as t → ∞, and interpret medically.
(d) Sketch the graph of A(t) for t ≥ 0, labelling the initial value, the value at t = 4, and the long-term asymptote. 7 marks Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — antiderivative 4 e^−0.5t + C using ∫ e^kx dx = (1/k) e^kx.

• 1 mark — uses A(0) = 4 to deduce C = 0 (so A(t) = 4 e^−0.5t).

Part (b) — 1 mark

• 1 mark — A(4) = 4 e⁻² ≈ 0.54 mg (accept 0.54 or exact 4/e²).

Part (c) — 2 marks

• 1 mark — states A(t) → 0 as t → ∞ with justification (e^−0.5t → 0).

• 1 mark — interprets medically: the tracer is essentially completely eliminated; A = 0 is the long-run baseline.

Part (d) — 2 marks

• 1 mark — correctly shaped curve: starts at (0, 4), strictly decreasing, concave up, approaching A = 0.

• 1 mark — labels: initial value 4 mg on the A-axis, the point (4, 0.54), and the horizontal asymptote A = 0.

Your response:

Stuck on the sketch? Plot only three points: (0, 4), (4, 0.54), and the asymptote A = 0 — that fixes the shape.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — ∫(3 e^2x − 5/x + 4x) dx (3 marks)

Sample response. ∫3 e^2x dx = (3/2) e^2x; ∫(−5/x) dx = −5 ln|x|; ∫4x dx = 2x². Combining: (3/2) e^2x − 5 ln|x| + 2x² + C.

Marking notes. 1 mark — correct (1/k) factor on the exponential ((3/2), not 3). 1 mark — correctly uses ln|x| for the 1/x term. 1 mark — combines with a single + C. Missing absolute value: −0.5; missing + C: −0.5.

1.2 — ∫(e^−x + x⁻¹ + x^1/2) dx (3 marks)

Sample response. ∫e^−x dx = −e^−x; ∫x⁻¹ dx = ln|x|; ∫x^1/2 dx = (2/3) x^3/2. Combining: −e^−x + ln|x| + (2/3) x^3/2 + C.

Marking notes. 1 mark — correct sign on −e^−x (k = −1 gives 1/k = −1). 1 mark — ln|x| (with absolute value). 1 mark — (2/3) x^3/2 from the power rule. A response writing ∫e^−x dx = e^−x (wrong sign) loses 1 mark.

1.3 — d/dx(e^kx/k) and two integrals (4 marks)

Sample response. By the chain rule, d/dx(e^kx/k) = (1/k) · k · e^kx = e^kx. Hence ∫e^kx dx = e^kx/k + C.
Applying with k = −3: ∫e^−3x dx = −(1/3) e^−3x + C.
Applying with k = 0.5 and coefficient 4: ∫4 e^0.5x dx = 4 · (1/0.5) e^0.5x + C = 8 e^0.5x + C.

Marking notes. 1 mark — correct derivative shown (with chain rule explicit). 1 mark — general formula ∫e^kx dx = e^kx/k + C stated or used. 1 mark each — the two specific integrals. Sign error on the k = −3 case is the most common slip.

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

(a) A(t) = ∫ (−2 e^−0.5t) dt = (−2) · (1/−0.5) · e^−0.5t + C = 4 e^−0.5t + C. [1 mark — antiderivative.]

Using A(0) = 4: 4 = 4 · 1 + C ⇒ C = 0, so A(t) = 4 e^−0.5t. [1 mark — constant fixed.]

(b) A(4) = 4 e⁻² = 4/e² ≈ 0.54 mg. [1 mark.]

(c) As t → ∞, e^−0.5t → 0, so A(t) → 4 · 0 = 0. [1 mark — limit with reason.]

Medically: the radioactive tracer is essentially completely eliminated from the patient's body in the long run; A = 0 is the asymptotic baseline that the body returns to. [1 mark — interpretation.]

(d) Sketch. A horizontal axis labelled t (hours), vertical axis A (mg). The curve starts at (0, 4) on the A-axis, decreases strictly, and is concave up; it passes through (4, 0.54) and approaches the horizontal asymptote A = 0 from above as t → ∞. [1 mark — correct shape; 1 mark — labels (0, 4), (4, 0.54), asymptote A = 0.]

A (mg)
 4 |*
   | \
   |  \
   |   \
   |    *_
   |       *_____ (4, 0.54)
   |             ____________
 0 |________________________________ A = 0  (asymptote)
   0      1      2      3      4      t (h)

Total: 7/7.

Band descriptors for marker.

Band 3: Completes (a) without finding C, or makes a sign error on the antiderivative. ≈ 1-2 marks.

Band 4: Completes (a) and (b); attempts (c) but only states the limit (no medical interpretation); no sketch or rough sketch only. ≈ 3-4 marks.

Band 5: All parts essentially correct; sketch present but missing one of the three labels. ≈ 5-6 marks.

Band 6: Full algebraic working, both the algebraic and physical interpretations in (c), and a sketch with all three labels (initial 4 mg, point (4, 0.54), asymptote A = 0). 7/7.