Source: CSSA Past Papers
Tags: integration, definite integrals, area between curves, area under a curve, exponential integration, volume of revolution, solids of revolution, logarithmic integration, inverse functions, parabolas, calculus, HSC, CSSA
These questions test your ability to set up and evaluate definite integrals for areas under curves and between curves, including exponentials and polynomials. Several require you to find unknown bounds, split regions, or rotate about the x-axis for volumes of revolution.
Definite integral
The integral of a function between two specific limits, giving a signed area. Written as ∫ from a to b of f(x) dx.
Area between curves
Found by integrating (top function − bottom function) over the interval where one curve lies above the other. Always take the absolute value of each piece if the curves cross.
Volume of revolution (disk method)
When a region is rotated about the x-axis, V = π ∫ from a to b of [f(x)]² dx. If the region is between two curves, subtract: V = π ∫ [f(x)]² − [g(x)]² dx.
Integrating with respect to y
When the boundary is easier to describe as x = f(y), integrate horizontally. Area = ∫ from y₁ to y₂ of x dy.
The area under y = eˣ from x = 0 to x = a is set equal to 3.
Set up the integral:
∫ from 0 to a of eˣ dx = [eˣ] from 0 to a = eᵃ − e⁰ = eᵃ − 1
Solve eᵃ − 1 = 3, so eᵃ = 4
Therefore a = ln 4 = 2 ln 2
Answer: (D) 2 ln 2
The trap here is forgetting to subtract 1 (the value at x = 0). If you forget, you get a = ln 3, which is option (C).
Part (i): Show the curves intersect at (−2, 8) and (2, 8)
Set 2x² = 12 − x²
3x² = 12, so x² = 4, giving x = ±2
When x = 2: y = 2(4) = 8. When x = −2: y = 2(4) = 8
Intersection points confirmed: (−2, 8) and (2, 8)
Part (ii): Area of the shaded region
Between x = −2 and x = 2, the upper curve is y = 12 − x² and the lower curve is y = 2x²
Area = ∫ from −2 to 2 of [(12 − x²) − 2x²] dx
= ∫ from −2 to 2 of (12 − 3x²) dx
The integrand is even (symmetric about x = 0), so this equals 2 × ∫ from 0 to 2 of (12 − 3x²) dx
= 2 × [12x − x³] from 0 to 2
= 2 × [(24 − 8) − 0]
= 2 × 16 = 32 units²
Answer: 32 units²
Part (i): Coordinates of T
T is where y = 2 − x² meets the x-axis, so 2 − x² = 0, giving x² = 2, x = √2 (taking the positive root from the diagram).
T = (√2, 0)
Part (ii): Volume of revolution
The shaded region lies between x = 0 and x = √2, bounded above by the two curves and below by the x-axis. The curves meet at R(1, 1).
Split into two sub-regions and rotate each about the x-axis:
From x = 0 to x = 1, the boundary curve above the x-axis is y = x³
From x = 1 to x = √2, the boundary curve above the x-axis is y = 2 − x²
Volume = π ∫ from 0 to 1 of (x³)² dx + π ∫ from 1 to √2 of (2 − x²)² dx
First integral:
π ∫ from 0 to 1 of x⁶ dx = π [x⁷/7] from 0 to 1 = π/7
Second integral:
Expand (2 − x²)² = 4 − 4x² + x⁴
π ∫ from 1 to √2 of (4 − 4x² + x⁴) dx = π [4x − 4x³/3 + x⁵/5] from 1 to √2
At x = √2: 4√2 − 4(2√2)/3 + (4√2)/5 = 4√2 − 8√2/3 + 4√2/5
Common denominator 15: (60√2 − 40√2 + 12√2)/15 = 32√2/15
At x = 1: 4 − 4/3 + 1/5 = (60 − 20 + 3)/15 = 43/15
Difference: 32√2/15 − 43/15
Total volume = π(1/7 + 32√2/15 − 43/15)
= π(15 + 7 × 32√2 − 7 × 43) / 105
= π(15 + 224√2 − 301) / 105
= π(224√2 − 286) / 105
Finding intersection points A and B
Set x³ − 2x = x², so x³ − x² − 2x = 0, giving x(x² − x − 2) = 0, so x(x − 2)(x + 1) = 0.
Intersections at x = −1, x = 0, x = 2 (points A, O, B respectively).
Area of the shaded region
The shaded region has two parts where different curves are on top:
From x = −1 to x = 0: compare the curves. At x = −0.5, x³ − 2x = 0.875 and x² = 0.25, so the cubic is above the parabola.
From x = 0 to x = 2: at x = 1, x³ − 2x = −1 and x² = 1, so the parabola is above the cubic.
Area = ∫ from −1 to 0 of [(x³ − 2x) − x²] dx + ∫ from 0 to 2 of [x² − (x³ − 2x)] dx
First integral: ∫ from −1 to 0 of (x³ − x² − 2x) dx
= [x⁴/4 − x³/3 − x²] from −1 to 0
= 0 − (1/4 + 1/3 − 1)
= 0 − (3/12 + 4/12 − 12/12) = 0 − (−5/12) = 5/12
Second integral: ∫ from 0 to 2 of (−x³ + x² + 2x) dx
= [−x⁴/4 + x³/3 + x²] from 0 to 2
= (−16/4 + 8/3 + 4) − 0
= −4 + 8/3 + 4 = 8/3
Total area = 5/12 + 8/3 = 5/12 + 32/12 = 37/12 ✓
This question asks for a horizontal area (integrating with respect to y).
From y = 2ˣ, invert to get x = log₂ y = ln y / ln 2
Area = ∫ from 1 to 4 of x dy = ∫ from 1 to 4 of (ln y / ln 2) dy = (1/ln 2) ∫ from 1 to 4 of ln y dy
Evaluate ∫ ln y dy using integration by parts:
∫ ln y dy = y ln y − y + C
So (1/ln 2) × [y ln y − y] from 1 to 4
At y = 4: 4 ln 4 − 4 = 4(2 ln 2) − 4 = 8 ln 2 − 4
At y = 1: 1(0) − 1 = −1
Difference: 8 ln 2 − 4 − (−1) = 8 ln 2 − 3
Area = (8 ln 2 − 3) / ln 2 = 8 − 3/ln 2
Answer: (B) 8 − 3/ln 2
Intersection at A (x = 3 given)
Verify: set 4x − x² = x² − 2x, giving 6x − 2x² = 0, so 2x(3 − x) = 0. Intersections at x = 0 and x = 3. Confirmed.
Area of the shaded region bounded by the two parabolas and the x-axis
Between x = 0 and x = 3, the curve y = 4x − x² is above y = x² − 2x. But y = x² − 2x dips below the x-axis (its roots are x = 0 and x = 2), so the "shaded region bounded by the two parabolas and the x-axis" needs care.
The region consists of:
From x = 0 to x = 2: the area between y = 4x − x² (above) and y = 0 (x-axis), since x² − 2x ≤ 0 in this interval. The shaded region here is the area under the upper parabola and above the x-axis.
From x = 2 to x = 3: the area between y = 4x − x² (above) and y = x² − 2x (below, now above the x-axis).
Area = ∫ from 0 to 2 of (4x − x²) dx + ∫ from 2 to 3 of [(4x − x²) − (x² − 2x)] dx
First integral:
= [2x² − x³/3] from 0 to 2 = 8 − 8/3 = 16/3
Second integral:
Integrand: 6x − 2x²
= [3x² − 2x³/3] from 2 to 3
At x = 3: 27 − 18 = 9
At x = 2: 12 − 16/3 = 20/3
Difference: 9 − 20/3 = 7/3
Total area = 16/3 + 7/3 = 23/3 units²
(Note: the exact shaded region depends on the diagram. An alternative reading, taking the full area between the two curves from 0 to 3, gives ∫ from 0 to 3 of (6x − 2x²) dx = [3x² − 2x³/3] from 0 to 3 = 27 − 18 = 9.)
Area under a curve: A = ∫ from a to b of f(x) dx
Area between two curves: A = ∫ from a to b of [f(x) − g(x)] dx, where f(x) ≥ g(x)
Volume of revolution about x-axis (disk): V = π ∫ from a to b of [f(x)]² dx
Integral of eˣ: ∫ eˣ dx = eˣ + C
Integral of ln y: ∫ ln y dy = y ln y − y + C (by parts)
Changing base: log₂ y = ln y / ln 2
⚠️ Always check which curve is on top before integrating. If they swap, split into separate integrals.
⚠️ When asked for the area bounded by curves and the x-axis, be careful: parts of a curve below the x-axis need separate treatment.
⚠️ For "area from y = ... to y = ..." questions, you must integrate with respect to y. Rewrite x as a function of y first.
⚠️ When solving for an unknown bound (like a in Q10), do not forget to subtract the lower-limit value of the antiderivative.
⚠️ Volumes of revolution often require splitting the region at intersection points.
Q: What is ∫ from 0 to a of eˣ dx, and what value of a makes this equal to 3?
A: The integral is eᵃ − 1. Setting eᵃ − 1 = 3 gives eᵃ = 4, so a = ln 4 = 2 ln 2.
Q: How do you find the area between y = 2x² and y = 12 − x² from x = −2 to x = 2?
A: Integrate (12 − x²) − 2x² = 12 − 3x² from −2 to 2. By symmetry this is 2 × ∫ from 0 to 2 of (12 − 3x²) dx = 32 units².
Q: When finding the area bounded by y = 2ˣ and the y-axis between y = 1 and y = 4, why do you integrate with respect to y?
A: Because the bounds are given as y-values and the region is measured horizontally from the y-axis to the curve. You rewrite x = ln y / ln 2 and integrate ∫ from 1 to 4 of (ln y / ln 2) dy.
Q: What is the general strategy when two curves cross within the interval of integration?
A: Find all intersection points within the interval, then split the integral at each crossing. On each sub-interval, determine which curve is on top and integrate (top − bottom). Sum the results.
definite integral, area under curve, area between curves, exponential function, eˣ integration, volume of revolution, disk method, washer method, solids of revolution, integration with respect to y, horizontal integration, logarithmic integration, parabola area, polynomial integration, CSSA past papers, HSC calculus, Texas A&M calculus