Grade 11 · Pure Mathematics 1 · Cambridge A-Level 9709 · Age 16–17
Welcome to Applications of Integration!
Now that you can perform integration, it is time to put it to work. This topic covers the most powerful geometric and physical uses of integration in Cambridge A-Level Pure Mathematics 1: computing areas bounded by curves, applying the trapezium rule as a numerical method, calculating volumes of solids of revolution, and using integration to solve kinematics problems involving displacement, velocity and acceleration.
Area = ∫[a,b] f(x) dx | V = π∫[a,b] y² dx | x = ∫v dt | Trap. Rule ≈ ½h[y₀ + 2(y₁+…+yₙ₋₁) + yₙ]
Learning Objectives
Calculate the area between a curve and the x-axis, handling regions below the axis correctly
Find the area between two curves by integrating the difference of the functions
Find the area between a curve and the y-axis by integrating x as a function of y
Apply the trapezium rule with n strips, and judge whether it over- or underestimates
Calculate the volume of a solid of revolution about the x-axis: V = π∫y² dx
Calculate volumes of revolution about the y-axis: V = π∫x² dy
Use integration to find displacement from velocity and velocity from acceleration
Distinguish clearly between distance travelled and displacement
Apply initial conditions to determine constants of integration in kinematics
Area Under Curve
∫f(x)dx; split at roots for regions below axis
Area Between Curves
∫(top − bottom)dx; find intersections first
Trapezium Rule
Numerical method; h=(b−a)/n; over/under estimates
Volumes of Revolution
V = π∫y²dx (x-axis); V = π∫x²dy (y-axis)
Kinematics
v = dx/dt; a = dv/dt; integrate to find position
Area — y-axis
Express x in terms of y; integrate from y=c to y=d
Distance vs Displacement
Split at sign changes of velocity; take absolute values
Initial Conditions
Use v(0) or x(0) to determine constant C after integrating
Learn 1 — Area Under a Curve
The Basic Area Formula
The area of the region bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is given by the definite integral. This assumes the curve stays above the x-axis throughout [a, b].
Area = ∫[a,b] f(x) dx (when f(x) ≥ 0 on [a, b])
Why the Integral Gives Area
Think of the region as infinitely many vertical strips of width dx and height f(x). Each strip has area f(x) dx. The integral sums all such strips continuously from x = a to x = b, giving the exact area. This is the geometric meaning of the definite integral.
When the Curve is Below the x-axis
When f(x) < 0 on [a, b], the definite integral ∫[a,b] f(x) dx gives a negative value. This is the signed area. The actual geometric area is the absolute value.
Key rule: If the question asks for "the area", it means the positive (geometric) area. You must identify any portions of the curve below the x-axis, handle them separately, and take absolute values.
Example: Find the area between y = x² − 4, the x-axis, x = 0 and x = 3.
Step 1: Find where y = 0: x² − 4 = 0 → x = ±2. Root at x = 2 in [0, 3]. Step 2: On [0, 2], y = x² − 4 < 0 (since y(1) = −3). On [2, 3], y > 0. Step 3: ∫[0,2] (x²−4) dx = [x³/3 − 4x]₀² = (8/3 − 8) − 0 = −16/3. Area = 16/3. Step 4: ∫[2,3] (x²−4) dx = [x³/3 − 4x]₂³ = (9 − 12) − (8/3 − 8) = −3 + 16/3 = 7/3. Total area = 16/3 + 7/3 = 23/3
Splitting Integrals at x-intercepts
Always find where the curve crosses the x-axis within the interval before computing area. Each section between consecutive roots needs its own integral, and you take the absolute value of each.
Total area = |∫[a,c] f(x) dx| + |∫[c,b] f(x) dx| where f(c) = 0
Area Between Curve and y-axis
To find the area between a curve and the y-axis (rather than the x-axis), integrate x as a function of y, with y as the variable of integration.
Area = ∫[c,d] x dy (where x is expressed as a function of y)
Example: Find the area between y = x², the y-axis, y = 0 and y = 4.
Express x in terms of y: x = √y = y^(1/2) (taking positive root since x ≥ 0).
Area = ∫[0,4] y^(1/2) dy = [2y^(3/2)/3]₀⁴ = 2·(4)^(3/2)/3 = 2·8/3 = 16/3
Example: Find the area between y = 2x − 1 and the y-axis from y = 1 to y = 5.
x = (y + 1)/2. Area = ∫[1,5] (y+1)/2 dy = [y²/4 + y/2]₁⁵ = (25/4 + 5/2) − (1/4 + 1/2) = 35/4 − 3/4 = 8
Always sketch the curve or at least check the sign of f(x) at a point in your interval before computing area. Write "the curve is above/below the x-axis on [a, b]" in your working — this shows the examiner you have thought about it. Splitting the integral correctly at roots is a guaranteed mark-winner.
Learn 2 — Area Between Two Curves
The Formula
If y = f(x) lies above y = g(x) on the interval [a, b] (meaning f(x) ≥ g(x) for all x in [a, b]), then the area of the region between the two curves is:
This works even when both curves are below the x-axis — as long as f is the top curve, f(x) − g(x) is positive and the integral gives the correct area directly.
Step-by-Step Method
Step 1: Find the intersection points of the two curves by solving f(x) = g(x). These give the limits a and b. Step 2: Determine which curve is on top in [a, b] by testing a value, e.g. check f(a + tiny bit) vs g(a + tiny bit). Step 3: Set up ∫[a,b] (top − bottom) dx and integrate. Step 4: Evaluate and state the answer as a positive number of square units.
Example 1 — Line and Parabola
Find the area enclosed between y = x + 2 and y = x².
Intersections: x + 2 = x² → x² − x − 2 = 0 → (x−2)(x+1) = 0 → x = −1, x = 2. Which is on top? At x = 0: line gives y = 2, parabola gives y = 0. Line is above. Area = ∫[−1,2] (x + 2 − x²) dx = [x²/2 + 2x − x³/3]₋₁²
At x = 2: 2 + 4 − 8/3 = 6 − 8/3 = 10/3.
At x = −1: 1/2 − 2 + 1/3 = −7/6.
Area = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2
Example 2 — Two Parabolas
Find the area between y = x² and y = 4 − x².
Intersections: x² = 4 − x² → 2x² = 4 → x² = 2 → x = ±√2. Which is on top? At x = 0: y = 4 − 0 = 4 vs y = 0. So y = 4 − x² is on top.
Area = ∫[−√2, √2] (4 − x² − x²) dx = ∫[−√2, √2] (4 − 2x²) dx
= [4x − 2x³/3]₋√2^√2 = 2·(4√2 − 2(√2)³/3) = 2(4√2 − 4√2/3) = 2·(8√2/3) = 16√2/3
Example 3 — Both Curves Above x-axis
Find the area between y = 4 − x² and y = x + 2.
Intersections: 4 − x² = x + 2 → x² + x − 2 = 0 → (x+2)(x−1) = 0 → x = −2, x = 1.
At x = 0: 4 − 0 = 4 vs 0 + 2 = 2. Top curve: y = 4 − x².
Area = ∫[−2,1] (4 − x² − x − 2) dx = ∫[−2,1] (2 − x − x²) dx
= [2x − x²/2 − x³/3]₋₂¹ = (2 − 1/2 − 1/3) − (−4 − 2 + 8/3)
= 7/6 − (−18/3 + 8/3) = 7/6 + 10/3 = 7/6 + 20/6 = 27/6 = 9/2
When Curves Swap Top and Bottom
If the two curves swap which one is on top within the interval (they cross again), you must split the integral at the interior crossing point and compute two separate area integrals. Never just integrate (f − g) across a region where the sign changes — that gives a net signed area, not a total geometric area.
The most common error is forgetting to find the intersection points and instead using incorrect limits. Always write "Set f(x) = g(x)" as the first step in finding area between curves. The intersection points are almost always the limits of integration.
Learn 3 — Trapezium Rule
What Is the Trapezium Rule?
Many functions are difficult or impossible to integrate analytically. The trapezium rule is a numerical method that approximates the area under a curve by replacing the curve with a series of straight-line segments, creating trapezoids whose areas we can easily sum.
where h = (b − a)/n is the width of each strip, n is the number of strips, and y₀, y₁, …, yₙ are the function values at the n + 1 equally spaced x-values.
The accuracy depends on the concavity of the curve:
Concave up (f''(x) > 0): The chord connecting each pair of points lies above the curve. Each trapezoid area is larger than the actual strip area. The rule overestimates the integral.
Concave down (f''(x) < 0): The chord lies below the curve. Each trapezoid area is smaller than the actual strip area. The rule underestimates the integral.
If the curve has both concave-up and concave-down sections, errors may partially cancel.
Memory aid:
Concave Up → Overestimate (like a smile ⌣ — chords go ABOVE the curve)
Concave Down → Underestimate (like a frown ⌢ — chords go BELOW the curve)
Improving Accuracy
The approximation improves as n increases (more strips, smaller h). Doubling n roughly quarters the error. In the limit n → ∞, the trapezium rule converges to the exact integral.
When applying the trapezium rule in an exam: (1) always set up a clear table of x and y values, (2) identify y₀ and yₙ clearly (used once), all other y values doubled (used twice), (3) state h = (b−a)/n explicitly. Missing the ½ factor or doubling the first/last y-values are the two most common errors. Always give your answer to the precision requested.
Learn 4 — Volumes of Revolution
What Is a Volume of Revolution?
When a curve y = f(x) is rotated through 360° (a full revolution) about the x-axis, it sweeps out a three-dimensional solid. We can find the volume of this solid by integrating — this is called the disc method, the only method required for Cambridge A-Level.
Volume About the x-axis
Each thin slice of the solid (at position x, width dx) is a disc with radius y and area πy². Its volume is πy² dx. Summing all discs gives:
V = π∫[a,b] y² dx
Substitute the expression for y in terms of x, then integrate. The answer has units of (length)³.
Step-by-Step Method (x-axis rotation)
Step 1: Write down V = π∫[a,b] y² dx. Step 2: Square y (the equation of the curve). Step 3: Integrate term by term. Step 4: Evaluate between limits and multiply by π. Step 5: State units and whether answer is exact (leave in terms of π if asked).
Example 1 — Rotate y = √x about x-axis
Find the volume of the solid formed when y = √x, x = 0 to x = 4, is rotated 360° about the x-axis.
y² = x
V = π∫[0,4] x dx = π[x²/2]₀⁴ = π(16/2 − 0) = 8π
Example 2 — Rotate y = x² + 1 about x-axis
Find the volume when y = x² + 1, from x = 0 to x = 2, is rotated about the x-axis.
When rotating about the y-axis, each horizontal slice is a disc of radius x. Integrate x² with respect to y.
V = π∫[c,d] x² dy (express x² as a function of y first)
Example 3 — Rotate y = x² about y-axis
Find the volume when y = x² (i.e. x = √y), from y = 0 to y = 9, is rotated about the y-axis.
x² = y
V = π∫[0,9] y dy = π[y²/2]₀⁹ = π(81/2) = 81π/2
Important Notes
• Cambridge 9709 only requires the disc method — no shells.
• Always write the π outside the integral as a factor.
• If the question says "leave your answer in terms of π", do not evaluate π numerically.
• Volume is always positive, so take care if the curve passes through negative y — but y² is always positive, so the integral is always non-negative.
The most common error is forgetting to square y before integrating. Writing V = π∫y dx (without squaring) is wrong. Always write y², expand it fully, then integrate term by term. Check by dimensions: if y is in cm, then y² is in cm², and π∫y²dx has units cm³ — a volume.
Learn 5 — Kinematics & Integration
The Kinematics Chain
In kinematics (the study of motion), three quantities are linked by differentiation and integration:
x (displacement) ⟷ v = dx/dt (velocity) ⟷ a = dv/dt = d²x/dt² (acceleration)
Differentiate to move right (displacement → velocity → acceleration). Integrate to move left (acceleration → velocity → displacement).
Finding Velocity from Acceleration
Given a(t), integrate to find v(t): v = ∫a dt + C
The constant C is found using an initial condition — typically the velocity at t = 0 (written v(0) = u, the initial speed).
Example: A particle has acceleration a = 6t − 2. Given v = 5 when t = 0, find v(t).
v = ∫(6t − 2) dt = 3t² − 2t + C
t = 0, v = 5: 5 = 0 − 0 + C → C = 5 v = 3t² − 2t + 5
Finding Displacement from Velocity
Given v(t), integrate to find x(t): x = ∫v dt + C
Use an initial condition x(0) = x₀ (initial position, often 0) to find C.
Example: A particle has velocity v = 2t² − 6t + 4. At t = 0 it is at the origin. Find the displacement at t = 3.
x = ∫(2t² − 6t + 4) dt = 2t³/3 − 3t² + 4t + C
x(0) = 0: C = 0.
x(3) = 2(27)/3 − 3(9) + 12 = 18 − 27 + 12 = 3 m
Distance Travelled ≠ Displacement
Displacement is the net change in position (can be positive, negative, or zero). Distance travelled is the total path length — always positive. They differ when the particle changes direction (v = 0 and changes sign).
Finding distance travelled when velocity changes sign:
Example: v = t² − 4t + 3 for 0 ≤ t ≤ 4. Find (a) displacement, (b) distance travelled.
(b) Distance: Find when v = 0: t² − 4t + 3 = 0 → (t−1)(t−3) = 0 → t = 1, t = 3.
v changes sign at t = 1 (positive to negative) and t = 3 (negative to positive).
∫[0,1] v dt = [t³/3 − 2t² + 3t]₀¹ = 1/3 − 2 + 3 = 4/3. (Positive — moving forward)
∫[1,3] v dt = [t³/3 − 2t² + 3t]₁³ = (9 − 18 + 9) − 4/3 = 0 − 4/3 = −4/3. Distance = 4/3.
∫[3,4] v dt = [t³/3 − 2t² + 3t]₃⁴ = 4/3 − 0 = 4/3. (Positive — moving forward) Total distance = 4/3 + 4/3 + 4/3 = 4 m
Setting Up from Given Information
Typical exam structure:
1. You are given either v(t) or a(t).
2. Integrate once (with initial condition) to find the next quantity.
3. Find when v = 0 to locate turning points / direction changes.
4. Use definite integrals (split if needed) for displacement or distance.
5. State clearly whether you've computed displacement or distance.
Always check whether the particle changes direction (v = 0) in the given time interval before calculating distance. If v never changes sign, then distance = |displacement|. Never confuse ∫[0,T] v dt (displacement) with ∫[0,T] |v| dt (distance). Cambridge mark schemes distinguish these clearly.
Worked Examples
Eight fully worked examples covering all key application topics for Cambridge A-Level 9709.
Example 1 — Area Under a Parabola
Q: Find the area enclosed between y = 6x − x² and the x-axis.
M1: Find roots of 6x − x² = 0: x(6 − x) = 0 → x = 0 or x = 6. Curve is above x-axis on (0, 6) since y(3) = 9 > 0.
M1: Area = ∫[0,6] (6x − x²) dx = [3x² − x³/3]₀⁶
M1: At x = 6: 3(36) − 216/3 = 108 − 72 = 36. At x = 0: 0.
A1: Area = 36 square units. 4 marks
Example 2 — Area Between a Line and a Curve
Q: Find the area enclosed between y = x² − 1 and y = 3.
M1: Intersections: x² − 1 = 3 → x² = 4 → x = ±2. Top curve: y = 3 (horizontal line is above parabola in [−2, 2]).
Example 3 — Area Below the x-axis (Split Integral)
Q: Find the total area enclosed between y = x³ − 4x, the x-axis, x = −2 and x = 2.
M1: Roots of x³ − 4x = x(x²−4) = x(x+2)(x−2) = 0: roots at x = −2, 0, 2. On [−2, 0]: y > 0 (check x = −1: −1+4 = 3). On [0, 2]: y < 0 (check x = 1: 1−4 = −3).
Q: The region bounded by y = x², the y-axis, y = 1 and y = 4 is rotated about the y-axis. Find the volume.
M1: Express x² in terms of y: y = x² → x² = y.
M1: V = π∫[1,4] x² dy = π∫[1,4] y dy = π[y²/2]₁⁴ = π(8 − 1/2) = π(15/2)
A1: Volume = 15π/2 cubic units. 3 marks
Example 7 — Kinematics: Finding Displacement from Velocity
Q: A particle moves so that its velocity at time t is v = 3t² − 12t + 9. Find the displacement from t = 0 to t = 4 and the total distance travelled.
M1: v = 0: 3t² − 12t + 9 = 0 → 3(t−1)(t−3) = 0 → t = 1, t = 3 (sign changes).
M1: Displacement = ∫[0,4] v dt = [t³ − 6t² + 9t]₀⁴ = (64−96+36) − 0 = 4 m.
M1: ∫[0,1] v dt = 1−6+9 = 4. ∫[1,3] v dt = 27−54+27−4 = −4. |−4| = 4. ∫[3,4] v dt = 4 − 0 = 4.
A1: Displacement = 4 m; total distance = 4 + 4 + 4 = 12 m. 5 marks
Example 8 — Kinematics: From Acceleration to Position
Q: A particle starts from rest at the origin. Its acceleration is a = 4 − 2t. Find its position at t = 5.
M1: v = ∫a dt = 4t − t² + C. v(0) = 0 → C = 0. v = 4t − t².
M1: x = ∫v dt = 2t² − t³/3 + C. x(0) = 0 → C = 0. x = 2t² − t³/3.
A1: x(5) = 2(25) − 125/3 = 50 − 125/3 = 150/3 − 125/3 = 25/3 m. 4 marks
Common Mistakes in Applications of Integration
These are the errors examiners see most in Cambridge 9709 scripts. Learn to recognise and avoid each one.
Mistake 1 — Not finding intersection points before area between curves
✗ Finding area between y = x² and y = x using limits x = 0 to x = 5 (wrong — these are not the natural boundaries)
✓ Always set f(x) = g(x) first. Here x² = x → x(x−1) = 0 → x = 0, x = 1. Use limits 0 to 1. Area = ∫[0,1] (x − x²) dx = 1/6.
Mistake 2 — Treating negative area as positive without splitting
✗ Area between y = sin x and x-axis from 0 to 2π = ∫[0,2π] sin x dx = [−cos x]₀^(2π) = (−1) − (−1) = 0
✓ The curve is above axis on [0,π] and below on [π,2π]. Split: |∫[0,π] sin x dx| + |∫[π,2π] sin x dx| = 2 + 2 = 4. The net integral is zero because positive and negative areas cancel.
Mistake 3 — Wrong trapezium rule formula (doubling first/last values)
✗ ½h[2(y₀ + y₁ + … + yₙ)] — treating all values the same
✓ The first (y₀) and last (yₙ) values are used ONCE. All interior values y₁, …, yₙ₋₁ are multiplied by 2. The formula is ½h[y₀ + 2(y₁ + … + yₙ₋₁) + yₙ].
Mistake 4 — Forgetting to square y in volume of revolution
✗ V = π∫[0,2] (x+1) dx (forgot to square the radius)
✓ V = π∫[0,2] (x+1)² dx. The disc radius is y = x+1, and disc area = πy², so you must square y before integrating. Forgetting to square is the #1 error in volume questions.
Mistake 5 — Not using initial conditions in kinematics
✗ Given a = 6t and v(0) = 3: writes v = 3t² (no constant)
✓ v = ∫6t dt = 3t² + C. Substitute t=0, v=3: 3 = 0 + C → C = 3. So v = 3t² + 3. Without the initial condition the constant C cannot be determined, giving the wrong velocity for all t > 0.
Mistake 6 — Confusing distance and displacement
✗ Particle with v = t² − 4: "distance from t=0 to t=3 = ∫[0,3] (t²−4) dt = [t³/3−4t]₀³ = 9−12 = −3 metres"
✓ v = 0 at t = 2. On [0,2]: v < 0; on [2,3]: v > 0. Distance = |∫[0,2] v dt| + ∫[2,3] v dt = 16/3 + 7/3 = 23/3 m. Displacement = −3 m (net). Distance is always positive.
Mistake 7 — Integrating y instead of y² for volume (y-axis rotation)
✗ For rotation about y-axis: V = π∫x dy (using x not x²)
✓ V = π∫[c,d] x² dy. Whether rotating about x-axis or y-axis, the formula always involves squaring the radius. For y-axis rotation the radius of each horizontal disc is x, so the formula is π∫x² dy.
Mistake 8 — Using wrong limits for trapezium rule (off by one strip)
✗ With n = 4 strips on [0, 2], using x-values 0, 0.5, 1, 1.5 (only 4 values, missing x = 2)
✓ n strips require n + 1 values: x₀, x₁, …, xₙ. For n = 4 on [0,2]: x = 0, 0.5, 1, 1.5, 2 — that's 5 values. Always count: number of y-values = number of strips + 1.
Key Formulas — Applications of Integration
Application
Formula
Key Notes
Area under curve (above x-axis)
Area = ∫[a,b] f(x) dx
f(x) ≥ 0 on [a,b]
Area under curve (below x-axis)
Area = |∫[a,b] f(x) dx|
Take absolute value of each section; split at x-intercepts
Area between curve and y-axis
Area = ∫[c,d] x dy
Express x as function of y; integrate w.r.t. y
Area between two curves
Area = ∫[a,b] [f(x)−g(x)] dx
f ≥ g on [a,b]; find intersections first for limits
Trapezium rule (n strips)
≈ ½h[y₀ + 2(y₁+…+yₙ₋₁) + yₙ]
h = (b−a)/n; n+1 y-values needed
Trapezium overestimates
when f''(x) > 0 (concave up)
Chords lie above the curve
Trapezium underestimates
when f''(x) < 0 (concave down)
Chords lie below the curve
Volume of revolution (x-axis)
V = π∫[a,b] y² dx
Square y first; multiply by π; exact answers leave π
Volume of revolution (y-axis)
V = π∫[c,d] x² dy
Express x² as function of y; integrate w.r.t. y
Velocity from displacement
v = dx/dt
Differentiate x w.r.t. t
Acceleration from velocity
a = dv/dt
Differentiate v w.r.t. t
Velocity from acceleration
v = ∫a dt + C
Use initial condition v(0) = u to find C
Displacement from velocity
x = ∫v dt + C
Use initial condition x(0) = x₀ to find C
Displacement (t₁ to t₂)
Δx = ∫[t₁,t₂] v dt
Can be negative (net change in position)
Distance travelled (t₁ to t₂)
d = ∫[t₁,t₂] |v| dt
Split at zeros of v; take absolute values of each part
Proof Bank
Two foundational proofs showing where the application formulas come from. Understanding these gives geometric insight that helps you avoid errors.
Proof 1 — Deriving the Trapezium Rule from Geometry
Divide [a, b] into n equal strips of width h = (b−a)/n. The x-values are x₀ = a, x₁ = a+h, x₂ = a+2h, …, xₙ = b.
Single trapezium: The strip between xᵢ₋₁ and xᵢ is bounded above by the chord joining (xᵢ₋₁, yᵢ₋₁) and (xᵢ, yᵢ). This shape is a trapezoid with parallel sides yᵢ₋₁ and yᵢ, and width h.
The first and last y-values appear once (they are each in only one trapezoid). Interior y-values appear in two adjacent trapezoids each, so they are doubled. Q.E.D.
Proof 2 — Volume of Revolution Formula from Riemann Sums (Disc Method)
Consider the curve y = f(x) rotated 360° about the x-axis from x = a to x = b.
Single disc: At position x, take a thin slice of thickness δx. When rotated, this slice sweeps out a disc (cylinder of negligible height) with:
— radius = y = f(x)
— cross-sectional area = πy² = π[f(x)]²
— volume = π[f(x)]² δx
Summing all discs (Riemann sum):
V ≈ Σ π[f(xᵢ)]² δx
Taking the limit as δx → 0 and the number of discs → ∞:
V = lim[δx→0] Σ π[f(x)]² δx = π∫[a,b] [f(x)]² dx = π∫[a,b] y² dx
This derivation shows that the formula V = π∫y²dx is simply the rigorous version of "sum up infinitely many thin disc volumes". The same argument with horizontal slices (radius = x, integrating over y) gives V = π∫x²dy for rotation about the y-axis. Q.E.D.
Area Between Curves Visualiser
Shows the shaded area between two curves. Use the toggle to switch between showing the area between y = x and y = x², or the single-curve area under y = x². The exact area is computed and displayed.
Mode: Area Between Curves
1.00
Area = 0
Pink = area between y=x and y=x²
Blue = single curve area (toggle)
Exact area shown
Exercise 1 — Area Under a Curve (10 Questions)
Exercise 2 — Area Between Two Curves (10 Questions)
Give answers as multiples of π (e.g. for 8π enter 8). Tolerance ±0.01 applied to the coefficient.
Exercise 5 — Kinematics with Integration (10 Questions)
Practice — 30 Mixed Questions
Challenge — 15 Harder Questions
Exam Style Questions
8 Cambridge-style multi-part questions with mark schemes. Attempt each before revealing.
Q1 [4 marks]
Find the area enclosed between the curve y = 4x − x² and the x-axis.
Roots: 4x − x² = x(4−x) = 0 → x = 0, x = 4. [M1]
Curve above x-axis on (0,4): y(2) = 4 > 0. [B1]
Area = ∫[0,4] (4x−x²) dx = [2x²−x³/3]₀⁴ = (32 − 64/3) − 0 = 32/3. [M1 A1] Area = 32/3 square units
Q2 [5 marks]
Find the area of the region enclosed between y = x² − 2x and y = 4 − x².
Intersections: x²−2x = 4−x² → 2x²−2x−4 = 0 → x²−x−2 = 0 → (x−2)(x+1) = 0 → x = −1, 2. [M1]
At x=0: y = 0 (lower) vs y = 4 (upper). Top curve: y = 4−x². [B1]
Area = ∫[−1,2] (4−x² − (x²−2x)) dx = ∫[−1,2] (4+2x−2x²) dx. [M1]
= [4x+x²−2x³/3]₋₁² = (8+4−16/3) − (−4+1+2/3) = (36/3−16/3) − (−9/3) = 20/3 + 3 = 29/3. [A1 A1] Area = 29/3 square units
Q3 [4 marks]
Use the trapezium rule with 5 strips to estimate ∫[0,1] (1 + x³) dx, giving your answer to 3 d.p. State whether your answer is an overestimate or underestimate, giving a reason.
h = 0.2. Values: y₀=1, y₁=1.008, y₂=1.064, y₃=1.216, y₄=1.512, y₅=2. [M1]
½×0.2×[1 + 2(1.008+1.064+1.216+1.512) + 2] = 0.1×[3 + 2(4.8)] = 0.1×[3+9.6] = 0.1×12.6 = 1.260. [M1 A1]
(Exact = [x+x⁴/4]₀¹ = 1.25. Rule gives 1.260 > 1.25.)
Overestimate because f''(x) = 6x ≥ 0 on [0,1], so curve is concave up and chords lie above the curve. [B1] ≈ 1.260; overestimate
Q4 [4 marks]
The region bounded by y = √(3x+1), the x-axis, x = 0 and x = 5 is rotated 360° about the x-axis. Find the exact volume.
A particle moves in a straight line with velocity v = 2t² − 8t + 6 m/s. (i) Find the times when the particle is at rest. (ii) Find the total distance travelled in the first 3 seconds.
(i) v = 0: 2t²−8t+6 = 0 → t²−4t+3 = 0 → (t−1)(t−3) = 0 → t = 1 s and t = 3 s. [M1 A1]
(ii) v < 0 on (1,3), v > 0 on (0,1) and (3,∞). Particle changes direction at t=1 and t=3. [M1]
∫[0,1] v dt = [2t³/3−4t²+6t]₀¹ = 2/3−4+6 = 8/3.
|∫[1,3] v dt| = |[2t³/3−4t²+6t]₁³| = |(18−36+18)−8/3| = |−8/3| = 8/3.
Total distance in [0,3] = 8/3 + 8/3 = 16/3 m. [M1 A1]
Q6 [5 marks]
The curve y = 1/x², the x-axis, x = 1 and x = 3 enclose a region R. Find (i) the area of R and (ii) the volume when R is rotated about the x-axis.
(i) Area = ∫[1,3] x⁻² dx = [−x⁻¹]₁³ = −1/3 − (−1) = 2/3. [M1 A1]
(ii) y² = x⁻⁴. V = π∫[1,3] x⁻⁴ dx = π[−x⁻³/3]₁³ = π(−1/81 + 1/3) = π(−1/81 + 27/81) = 26π/81. [M1 M1 A1] V = 26π/81 cubic units
Q7 [5 marks]
A particle starts from rest at a point O and moves along a straight line. Its acceleration after t seconds is a = 6 − 2t m/s². (i) Find the velocity when t = 4. (ii) Find the displacement from O when t = 6.
(i) v = ∫(6−2t) dt = 6t − t² + C. v(0) = 0 → C = 0. v = 6t − t². [M1]
v(4) = 24 − 16 = 8 m/s. [A1]
(ii) x = ∫(6t−t²) dt = 3t²−t³/3 + C. x(0) = 0 → C = 0. x = 3t²−t³/3. [M1]
x(6) = 3(36) − 216/3 = 108 − 72 = 36 m. [M1 A1]
Q8 [6 marks]
The region bounded by the curve y = x(2−x) and the line y = x is S. (i) Find the area of S. (ii) Find the volume of the solid formed when S is rotated through 360° about the x-axis.
(i) Intersections: x(2−x) = x → 2x−x² = x → x−x² = 0 → x(1−x) = 0 → x = 0, x = 1. [M1]
At x = 0.5: curve y = 0.75, line y = 0.5. Curve above on (0,1). [B1]
Area = ∫[0,1] (x(2−x) − x) dx = ∫[0,1] (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6. [M1 A1]
(ii) V = π∫[0,1] [(x(2−x))² − x²] dx = π∫[0,1] [x²(2−x)² − x²] dx = π∫[0,1] x²[(2−x)²−1] dx. [M1]
(2−x)²−1 = 4−4x+x²−1 = 3−4x+x². So x²(3−4x+x²) = 3x²−4x³+x⁴.
V = π[x³ − x⁴ + x⁵/5]₀¹ = π(1−1+1/5) = π/5. [A1]
Past Paper Questions
5 questions drawn from Cambridge A-Level 9709 Pure 1 past papers on Applications of Integration. Attempt each before revealing the solution.
Past Paper Q1 — 9709/11/O/N/21 Q8 [6 marks]
The curve y = 2x^(1/2) and the line y = x intersect at O and P. Find P. Find the area enclosed between the curve and the line.
Intersection: 2x^(1/2) = x → 4x = x² → x(x−4) = 0 → x = 0 or x = 4. [M1]
At x = 4: y = 2(2) = 4. P = (4, 4). [A1]
On (0,4): curve above line (check x=1: y=2 vs y=1). [B1]
Area = ∫[0,4] (2x^(1/2) − x) dx = [4x^(3/2)/3 − x²/2]₀⁴ [M1]
= (4·8/3 − 8) − 0 = 32/3 − 8 = 32/3 − 24/3 = 8/3 square units. [M1 A1]
Past Paper Q2 — 9709/12/M/J/20 Q9 [7 marks]
The region R is bounded by the curves y = x² and y = 4x − x². Find the area of R. Find the volume when R is rotated through 360° about the x-axis.
Intersections: x² = 4x−x² → 2x²−4x = 0 → 2x(x−2) = 0 → x = 0, x = 2. [M1]
Top curve on (0,2): y = 4x−x² (check x=1: 3 vs 1). [B1]
Area = ∫[0,2] (4x−2x²) dx = [2x²−2x³/3]₀² = (8−16/3) = 8/3. [M1 A1]
V = π∫[0,2] [(4x−x²)² − (x²)²] dx = π∫[0,2] [16x²−8x³+x⁴−x⁴] dx = π∫[0,2] (16x²−8x³) dx. [M1]
= π[16x³/3−2x⁴]₀² = π(128/3 − 32) = π(128/3 − 96/3) = 32π/3. [M1 A1]
Past Paper Q3 — 9709/13/O/N/19 Q7 [6 marks]
Use the trapezium rule with ordinates at x = 1, 2, 3, 4, 5 to estimate ∫[1,5] ln x dx. Given that the exact value is (5 ln 5 − 4), find the percentage error in your approximation.
A particle P starts from rest at a point O and moves in a straight line. Its velocity v m/s at time t s is given by v = 3t² − 12t + 9. Find (i) the values of t when P is instantaneously at rest, (ii) the displacement of P from O when t = 4, (iii) the total distance travelled by P in the first 4 seconds.
(i) 3t²−12t+9 = 0 → t²−4t+3 = 0 → (t−1)(t−3) = 0 → t = 1 and t = 3. [M1 A1]
(ii) x = ∫v dt = t³−6t²+9t + C. x(0) = 0 → C = 0. [M1]
x(4) = 64−96+36 = 4 m. [A1]
(iii) On [0,1]: v > 0. On [1,3]: v < 0. On [3,4]: v > 0. [M1]
x(1) = 1−6+9 = 4. x(3) = 27−54+27 = 0. x(4) = 4.
Distances: 0→1: |4−0|=4. 1→3: |0−4|=4. 3→4: |4−0|=4. Total distance = 12 m. [M1 A1]
Past Paper Q5 — 9709/12/O/N/17 Q11 [8 marks]
The curve C has equation y = (x−1)√x. (i) Find the x-coordinates where C crosses the x-axis. (ii) Find the area enclosed between C and the x-axis. (iii) Find the volume when this region is rotated through 360° about the x-axis.
(i) (x−1)√x = 0 → x = 0 or x = 1. C crosses at x = 0 and x = 1. [B1]
(ii) On [0,1]: y = (x−1)√x. Check x=0.25: (−0.75)(0.5) = −0.375 < 0. Curve below axis. [B1]
y = x^(3/2) − x^(1/2). ∫y dx = 2x^(5/2)/5 − 2x^(3/2)/3. [M1 A1]
∫[0,1] y dx = 2/5 − 2/3 = 6/15 − 10/15 = −4/15. Area = 4/15. [A1]
(iii) y² = (x−1)²x = x³ − 2x² + x. [M1]
V = π∫[0,1] (x³−2x²+x) dx = π[x⁴/4 − 2x³/3 + x²/2]₀¹ = π(1/4−2/3+1/2) = π(3/12−8/12+6/12) = π(1/12). [M1 A1] V = π/12