Applications of Integration

This lesson covers the practical applications of integration — particularly in kinematics and modelling — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to be able to find displacement from velocity, velocity from acceleration, and apply integration to real-world modelling problems.

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Integration in Kinematics

Kinematics is the study of motion. The three key quantities are:

• Displacement s (position, measured from a reference point)
• Velocity v (rate of change of displacement, v = ds/dt)
• Acceleration a (rate of change of velocity, a = dv/dt)

The Relationships

Differentiation goes "down" this chain:

• s → v = ds/dt
• v → a = dv/dt

Integration goes "up" this chain:

• a → v = ∫ a dt
• v → s = ∫ v dt

To find	You integrate	Formula
Velocity from acceleration	∫ a dt	v = ∫ a dt
Displacement from velocity	∫ v dt	s = ∫ v dt
Distance travelled over [t₁, t₂]	∫ from t₁ to t₂ of	v

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Finding Velocity from Acceleration

If the acceleration is given as a function of time, integrate to find velocity.

Example 1

A particle moves in a straight line with acceleration a = 6t - 2 m/s². At t = 0, the velocity is 5 m/s. Find v in terms of t.

Step 1: Integrate acceleration. v = ∫ (6t - 2) dt = 3t² - 2t + c

Step 2: Use the initial condition v = 5 when t = 0. 5 = 0 - 0 + c → c = 5

Step 3: v = 3t² - 2t + 5

Example 2

A particle has acceleration a = 10e^(-2t) m/s² and starts from rest (v = 0 at t = 0). Find v.

v = ∫ 10e^(-2t) dt = -5e^(-2t) + c

When t = 0: 0 = -5(1) + c → c = 5

v = 5 - 5e^(-2t) = 5(1 - e^(-2t))

Note: As t → ∞, v → 5 m/s (the velocity approaches a limiting value).

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Finding Displacement from Velocity

If velocity is given as a function of time, integrate to find displacement.

Example 1

A particle moves with velocity v = 4t - t² m/s. At t = 0, the displacement from the origin is 2 m. Find s in terms of t.

s = ∫ (4t - t²) dt = 2t² - t³/3 + c

When t = 0: 2 = 0 + c → c = 2

s = 2t² - t³/3 + 2

Example 2

A particle has velocity v = 3 sin(2t) m/s and is at the origin when t = 0. Find s.

s = ∫ 3 sin(2t) dt = -(3/2) cos(2t) + c

When t = 0: 0 = -(3/2)(1) + c → c = 3/2

s = (3/2)(1 - cos(2t))

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Distance vs Displacement

Displacement is the position relative to a reference point — it can be positive or negative.

Distance is the total length of the path travelled — it is always positive.

The Critical Difference

If a particle moves forward and then backward:

• The displacement may be small (or even zero) because the backward motion cancels the forward motion.
• The distance is the sum of all the lengths travelled, regardless of direction.

Finding Total Distance

To find the total distance travelled between t = t₁ and t = t₂:

1. Find when v = 0 (the particle changes direction).
2. Calculate the displacement for each interval separately.
3. Take the absolute value of each displacement and add them up.

Example: A particle moves with velocity v = t² - 4t + 3 m/s. Find the total distance travelled from t = 0 to t = 4.

Step 1: Find when v = 0. t² - 4t + 3 = 0 → (t - 1)(t - 3) = 0 → t = 1 or t = 3

Step 2: Check the sign of v in each interval.

• 0 ≤ t ≤ 1: v(0.5) = 0.25 - 2 + 3 = 1.25 > 0 (moving forward)
• 1 ≤ t ≤ 3: v(2) = 4 - 8 + 3 = -1 < 0 (moving backward)
• 3 ≤ t ≤ 4: v(3.5) = 12.25 - 14 + 3 = 1.25 > 0 (moving forward)

Step 3: Integrate each part.

Interval [0, 1]: ∫ from 0 to 1 of (t² - 4t + 3) dt = [t³/3 - 2t² + 3t] from 0 to 1 = 1/3 - 2 + 3 = 4/3

Interval [1, 3]: ∫ from 1 to 3 of (t² - 4t + 3) dt = [t³/3 - 2t² + 3t] from 1 to 3 = (9 - 18 + 9) - (1/3 - 2 + 3) = 0 - 4/3 = -4/3

Interval [3, 4]: ∫ from 3 to 4 of (t² - 4t + 3) dt = [t³/3 - 2t² + 3t] from 3 to 4 = (64/3 - 32 + 12) - 0 = 64/3 - 20 = 4/3

Step 4: Total distance = |4/3| + |-4/3| + |4/3| = 4/3 + 4/3 + 4/3 = 4

Exam Tip: If the question asks for "distance", you must account for changes of direction. If it asks for "displacement", you can just evaluate a single integral.

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Variable Acceleration Problems

At A-Level, many kinematics problems involve non-constant (variable) acceleration. The standard SUVAT equations only work for constant acceleration, so when acceleration varies, you must use calculus.

Complete Worked Example

A particle P moves along the x-axis. At time t seconds (t ≥ 0), the velocity of P is v = (2t - 1)(t - 3) m/s. At t = 0, P is at the origin.

(a) Find the acceleration when t = 2.

Expand v: v = 2t² - 7t + 3

a = dv/dt = 4t - 7

At t = 2: a = 8 - 7 = 1 m/s²

(b) Find the displacement of P from the origin when t = 3.

s = ∫ from 0 to 3 of (2t² - 7t + 3) dt = [2t³/3 - 7t²/2 + 3t] from 0 to 3

= (18 - 63/2 + 9) - 0

= 27 - 63/2

= 54/2 - 63/2

= -9/2

So the displacement is -4.5 m (P is 4.5 m in the negative direction from the origin).

(c) Find the total distance travelled from t = 0 to t = 3.

Find when v = 0: (2t - 1)(t - 3) = 0 → t = 0.5 or t = 3

Check signs: v(0) = (-1)(-3) = 3 > 0 and v(1) = (1)(-2) = -2 < 0

Interval [0, 0.5]: ∫ = [2t³/3 - 7t²/2 + 3t] from 0 to 0.5 = 2(0.125)/3 - 7(0.25)/2 + 1.5 = 1/12 - 7/8 + 3/2 = 2/24 - 21/24 + 36/24 = 17/24

Interval [0.5, 3]: displacement = -9/2 - 17/24 = -108/24 - 17/24 = -125/24

Total distance = 17/24 + 125/24 = 142/24 = 71/12 ≈ 5.917 m

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Modelling with Integration

Integration is used to model many real-world situations beyond kinematics.

Example: Rate of Flow

Water flows into a tank at a rate of dV/dt = 100 - 2t litres per minute, where t is the time in minutes. Find the total volume of water that flows into the tank during the first 30 minutes.

V = ∫ from 0 to 30 of (100 - 2t) dt = [100t - t²] from 0 to 30 = (3000 - 900) - 0 = 2100 litres

Example: Marginal Cost (Economics)

The marginal cost of producing x items is dC/dx = 20 + 0.4x pounds. The fixed cost is £500 (when x = 0, C = 500). Find the total cost function.

C = ∫ (20 + 0.4x) dx = 20x + 0.2x² + c

When x = 0: 500 = c → c = 500

C = 0.2x² + 20x + 500

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Integration with Parametric Equations

If a curve is given in parametric form x = f(t), y = g(t), the area under the curve can be found using:

Area = ∫ y dx = ∫ y (dx/dt) dt

Example: A curve is defined by x = t², y = t³ for 0 ≤ t ≤ 2. Find the area under the curve.

dx/dt = 2t

Area = ∫ from 0 to 2 of t³ × 2t dt = ∫ from 0 to 2 of 2t⁴ dt = [2t⁵/5] from 0 to 2 = 64/5 - 0 = 64/5

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Common Mistakes

1. Confusing distance and displacement — Distance is always positive; displacement can be negative.
2. Forgetting initial conditions — Always use given values of v or s at t = 0 to find the constant.
3. Using SUVAT with variable acceleration — SUVAT only works for constant acceleration. For variable acceleration, you must integrate.
4. Not finding when v = 0 — For distance problems, you must identify when the particle changes direction.
5. Mixing up the chain — Integrate acceleration to get velocity, integrate velocity to get displacement. Do not skip steps.

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Summary

• Velocity from acceleration: v = ∫ a dt.
• Displacement from velocity: s = ∫ v dt.
• For total distance, find when v = 0 (direction changes), calculate each interval separately, and add the absolute values.
• Use initial/boundary conditions to find the constant of integration.
• Integration applies to many modelling contexts: rates of flow, costs, parametric curves, and more.
• SUVAT equations do NOT apply when acceleration varies — use calculus instead.

A-Level Deep Dive: Applications of Integration

Spec mapping

Edexcel 9MA0-02 specification section 10 — Integration covers $x^n$xn (excluding $n = -1$n=−1), and related sums, differences and constant multiples; integrate $e^{kx}$ekx, $1/x$1/x, $\sin kx$sinkx, $\cos kx$coskx and related functions; evaluate definite integrals; use the integral as a sum, including for finding the area under a curve and the area between two curves; use integration to find the mean value of a function over an interval (refer to the official specification document for exact wording). Although integration is presented as the final pure topic, it is examined synoptically across section 8 (Differentiation, via the Fundamental Theorem), section 4 (Differential equations, via separation of variables), 9MA0-03 Mechanics (where $\int v\,dt = s$∫vdt=s and $\int a\,dt = v$∫adt=v) and 9MA0-03 Statistics (where the integral of a probability density function over an interval gives a probability). The Edexcel formula booklet lists standard integrals ($\int e^{kx}\,dx$∫ekxdx, $\int \sec^2 kx\,dx$∫sec2kxdx, etc.) but does not list the mean-value formula $\bar{f} = \dfrac{1}{b-a}\int_a^b f(x)\,dx$fˉ​=b−a1​∫ab​f(x)dx — this must be memorised.

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Worked example with full mark scheme

Question (8 marks): The curve $C$C has equation $y = 4 - x^2$y=4−x2 for $0 \leq x \leq 2$0≤x≤2. The region $R$R is bounded by $C$C, the $x$x-axis and the $y$y-axis.

(a) Find the exact area of $R$R. (3)

(b) Find the mean value of $y$y over the interval $0 \leq x \leq 2$0≤x≤2. (3)

(c) Hence state the height of the rectangle, on base $0 \leq x \leq 2$0≤x≤2, with the same area as $R$R. (2)

Solution with mark scheme:

(a) Step 1 — set up the definite integral.

$\text{Area} = \int_0^2 (4 - x^2)\,dx$Area=∫02​(4−x2)dx

M1 — correct integral with correct limits and integrand. Common error: integrating $|4 - x^2|$∣4−x2∣ unnecessarily — on $[0, 2]$[0,2] the curve sits above the $x$x-axis, so no modulus is required.

Step 2 — integrate.

$\int_0^2 (4 - x^2)\,dx = \left[4x - \dfrac{x^3}{3}\right]_0^2 = \left(8 - \dfrac{8}{3}\right) - 0 = \dfrac{24 - 8}{3} = \dfrac{16}{3}$∫02​(4−x2)dx=[4x−3x3​]02​=(8−38​)−0=324−8​=316​

M1 — correct antiderivative $4x - x^3/3$4x−x3/3.

A1 — exact area $\dfrac{16}{3}$316​.

(b) Step 1 — apply the mean-value formula.

$\bar{y} = \dfrac{1}{2 - 0}\int_0^2 (4 - x^2)\,dx$yˉ​=2−01​∫02​(4−x2)dx

M1 — correct mean-value formula with correct limits.

Step 2 — substitute the area from (a).

$\bar{y} = \dfrac{1}{2} \cdot \dfrac{16}{3} = \dfrac{8}{3}$yˉ​=21​⋅316​=38​

M1 — correct simplification using the previous result (the "Hence" command in spirit, even if not stated).

A1 — exact mean value $\dfrac{8}{3}$38​.

(c) Step 1 — interpret the mean value geometrically.

The mean value is the height of the rectangle, on the same base, with area equal to that of $R$R. So the rectangle has base $2$2 and height $\dfrac{8}{3}$38​, giving area $2 \cdot \dfrac{8}{3} = \dfrac{16}{3}$2⋅38​=316​, matching part (a).

B1 — stating the height is $\dfrac{8}{3}$38​.

B1 — verifying area $2 \cdot \bar{y} = \dfrac{16}{3}$2⋅yˉ​=316​, confirming consistency with (a).

Total: 8 marks (M3 A2 B2 with one further A reserved for presentation discipline; effectively M3 A3 B2 in the Edexcel split).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): A particle moves in a straight line so that its velocity at time $t$t seconds ($t \geq 0$t≥0) is $v = 3t^2 - 12t + 9$v=3t2−12t+9 m s$^{-1}$−1.

(a) Find the times at which the particle is instantaneously at rest. (2)

(b) Find the total distance travelled by the particle in the first 4 seconds, justifying your method. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — setting $3t^2 - 12t + 9 = 0$3t2−12t+9=0 and factorising: $3(t - 1)(t - 3) = 0$3(t−1)(t−3)=0.
• A1 (AO1.1b) — $t = 1$t=1 and $t = 3$t=3.

(b)

• M1 (AO3.1a) — recognising that "total distance" requires splitting the interval at sign changes of $v$v, not simply evaluating $\int_0^4 v\,dt$∫04​vdt (which would give net displacement).
• M1 (AO1.1b) — splitting into $\int_0^1 v\,dt$∫01​vdt, $\int_1^3 v\,dt$∫13​vdt, $\int_3^4 v\,dt$∫34​vdt and taking absolute values.
• A1 (AO1.1b) — evaluating: $\int_0^1 v\,dt = [t^3 - 6t^2 + 9t]_0^1 = 4$∫01​vdt=[t3−6t2+9t]01​=4; $\int_1^3 v\,dt = [t^3 - 6t^2 + 9t]_1^3 = 0 - 4 = -4$∫13​vdt=[t3−6t2+9t]13​=0−4=−4; $\int_3^4 v\,dt = [t^3 - 6t^2 + 9t]_3^4 = 4 - 0 = 4$∫34​vdt=[t3−6t2+9t]34​=4−0=4.
• A1 (AO2.5) — total distance $= 4 + 4 + 4 = 12$=4+4+4=12 m, with explicit statement that the negative integral is reversed because the particle moves backwards on $1 < t < 3$1<t<3.

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 1, with the remaining mark embedded in AO1. This is an AO3-heavy question — Edexcel uses mechanics-flavoured integration to test whether candidates recognise when net displacement and total distance differ. Candidates who simply compute $\int_0^4 v\,dt = 4$∫04​vdt=4 score the M-marks for technique but lose the A-marks for misinterpretation.

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Synoptic links

Connects to:

1. All integration techniques (sections 10.1–10.6): applications draw on every method — power rule ($\int x^n\,dx$∫xndx), exponential / logarithmic ($\int e^{kx}\,dx$∫ekxdx, $\int 1/x\,dx$∫1/xdx), trigonometric ($\int \sin kx\,dx$∫sinkxdx), substitution and integration by parts. A typical Paper 2 application question fixes the interpretation (area, volume, mean) and forces the candidate to choose the right technique given the integrand.