Source: ACE Exam Paper 1 ADV
Tags: integration, definite integrals, area between curves, trapezoidal rule, antiderivative, displacement, velocity, acceleration, motion, sec squared
Integration is differentiation in reverse, and it is tested both as a pure technique (evaluate the integral) and as an applied tool (area between curves, displacement from velocity, trapezoidal approximation). The exam expects fluency with standard forms, comfort with substitution by inspection, and the ability to set up area integrals from graphs.
Antiderivative / indefinite integral
A function F(x) whose derivative is f(x). Written ∫f(x) dx = F(x) + C.
Definite integral
∫ from a to b of f(x) dx = F(b) - F(a). Gives the signed area under the curve between x = a and x = b.
Area between curves
∫ from a to b of [upper function - lower function] dx. Always take the absolute value or arrange so the integrand is positive.
Trapezoidal rule
An approximation for a definite integral using trapezoids. Formula: ∫ ≈ (h/2)[f(x₀) + f(xₙ) + 2(f(x₁) + f(x₂) + ... + f(xₙ₋₁))], where h is the width of each strip.
Displacement
The integral of velocity with respect to time: x = ∫v dt. Gives position (with a constant of integration determined by initial conditions).
Acceleration
The derivative of velocity with respect to time: a = dv/dt.
Split into two integrals:
∫₀¹ e³ˣ dx = (1/3)e³ˣ evaluated from 0 to 1 = (1/3)e³ - 1/3
∫₀¹ 1 dx = 1
Total = (1/3)e³ - 1/3 + 1 = (1/3)e³ + 2/3 = (1/3)(e³ + 2)
The answer is (D).
∫(2x + 3)⁻⁴ dx uses the reverse chain rule.
The inner function is (2x + 3), whose derivative is 2
∫(2x + 3)⁻⁴ dx = (1/2) × (2x + 3)⁻³ / (-3) + C = -1 / [6(2x + 3)³] + C
The general pattern: ∫(ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ / [a(n + 1)] + C, provided n ≠ -1.
∫₀^{π/8} sec²(2x) dx:
The antiderivative of sec²(2x) is (1/2) tan(2x)
Evaluate from 0 to π/8: (1/2) tan(π/4) - (1/2) tan(0) = (1/2)(1) - 0 = 1/2
For y = x² + 1 and y = 3x + 1, find where they intersect:
x² + 1 = 3x + 1, so x² - 3x = 0, giving x(x - 3) = 0
Intersection at x = 0 and x = 3
Between x = 0 and x = 3, the line y = 3x + 1 is above the parabola y = x² + 1.
Area = ∫₀³ [(3x + 1) - (x² + 1)] dx = ∫₀³ (3x - x²) dx
= [3x²/2 - x³/3] from 0 to 3
= (27/2 - 9) - 0
= 27/2 - 9 = 9/2 = 4.5 square units
Antiderivative: x³/6 + x
At x = 2: 8/6 + 2 = 4/3 + 2 = 10/3
At x = -1: -1/6 + (-1) = -7/6
Result: 10/3 - (-7/6) = 10/3 + 7/6 = 20/6 + 7/6 = 27/6 = 9/2
Let u = 3x + 4, so du = 3 dx.
When x = -1: u = 1. When x = 4: u = 16.
∫₁¹⁶ u^{1/2} × (1/3) du = (1/3) × [u^{3/2} / (3/2)] from 1 to 16
= (1/3) × (2/3) × [16^{3/2} - 1^{3/2}]
= (2/9) × (64 - 1) = (2/9)(63) = 14
Given five y-values at equal spacing h = 1.5 m: y₀ = 0, y₁ = 4.5, y₂ = 5.1, y₃ = 3.6, y₄ = 0. (The boundary starts and ends at ground level, with heights measured at intervals along the base.)
Area ≈ (h/2)[y₀ + y₄ + 2(y₁ + y₂ + y₃)] = (1.5/2)[0 + 0 + 2(4.5 + 5.1 + 3.6)] = 0.75 × 2(13.2) = 0.75 × 26.4 = 19.8 m²
For the rain calculation: 25 mm = 0.025 m of rain over 19.8 m².
Volume = 19.8 × 0.025 = 0.495 m³
1 m³ = 1000 L, so volume = 495 litres
Given v = 1 - 2 sin 2t, starting at the origin (x = 0 when t = 0).
(a) Displacement as a function of t:
x = ∫v dt = ∫(1 - 2 sin 2t) dt = t + cos 2t + C
At t = 0, x = 0: 0 = 0 + cos 0 + C = 1 + C, so C = -1.
x = t + cos 2t - 1
(b) Position at t = π/3:
x = π/3 + cos(2π/3) - 1 = π/3 + (-1/2) - 1 = π/3 - 3/2
(c) Acceleration:
a = dv/dt = d/dt(1 - 2 sin 2t) = -4 cos 2t
(d) Graph of acceleration for 0 ≤ t ≤ π:
a = -4 cos 2t is a cosine curve reflected vertically, with amplitude 4 and period π. It starts at a(0) = -4, crosses zero at t = π/4, reaches a maximum of 4 at t = π/2, crosses zero again at t = 3π/4, and returns to -4 at t = π.
(e) Maximum acceleration:
The maximum value of -4 cos 2t is 4, occurring when cos 2t = -1, i.e. at t = π/2.
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C, n ≠ -1
∫eᵃˣ dx = (1/a)eᵃˣ + C
∫sin(ax) dx = -(1/a)cos(ax) + C
∫cos(ax) dx = (1/a)sin(ax) + C
∫sec²(ax) dx = (1/a)tan(ax) + C
∫(ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ / [a(n + 1)] + C
Trapezoidal rule: ∫ ≈ (h/2)[f(x₀) + f(xₙ) + 2(f(x₁) + ... + f(xₙ₋₁))]
Displacement: x = ∫v dt
Velocity: v = dx/dt
Acceleration: a = dv/dt
⚠️ When integrating eᵃˣ, do not forget the 1/a factor. This is the single most common slip in integration questions.
⚠️ For area between curves, always check which function is on top in the given interval. If you get a negative area, your subtraction is backwards.
⚠️ In the trapezoidal rule, the first and last y-values are counted once; every interior value is counted twice. Mixing this up halves your accuracy.
⚠️ Motion questions always state initial conditions. Use them to find C. If the object starts at the origin, x(0) = 0.
⚠️ For ∫₀^{π/8} sec²(2x) dx, remember sec² integrates to tan, not the other way around. And do not forget the coefficient correction for the inner 2x.
Q: Evaluate ∫₀¹ (e³ˣ + 1) dx.
A: (1/3)(e³ + 2)
Q: Find ∫(2x + 3)⁻⁴ dx.
A: -1/[6(2x + 3)³] + C
Q: Find the area enclosed between y = x² + 1 and y = 3x + 1.
A: Intersect at x = 0 and x = 3. Area = ∫₀³ (3x - x²) dx = 9/2 = 4.5 square units.
Q: Use the trapezoidal rule with four intervals and h = 1.5 to approximate an area, given boundary heights 0, 4.5, 5.1, 3.6, 0.
A: Area ≈ (1.5/2)[0 + 0 + 2(4.5 + 5.1 + 3.6)] = 19.8 m².
Q: Given v = 1 - 2 sin 2t and x(0) = 0, find the displacement function.
A: x = t + cos 2t - 1.
Q: What is the maximum acceleration if a = -4 cos 2t?
A: The maximum value is 4 (when cos 2t = -1).
integration, definite integral, indefinite integral, antiderivative, area under curve, area between curves, trapezoidal rule, trapezoid approximation, numerical integration, reverse chain rule, displacement, velocity, acceleration, motion, kinematics, sec squared, HSC maths advanced, Year 12 integration