Practice Maths

Applications of Integration

Key Terms

Displacement
= ∫ab v(t) dt   (can be negative — net change in position)
Total distance
= ∫ab |v(t)| dt   (always positive — split where v(t) changes sign)
Total change
from a rate: if dQ/dt is a rate, then ∫ab dQ/dt dt = Q(b) − Q(a)
Average value
of f on [a,b]: f̄ = (1/(b−a)) ∫ab f(x) dx
Initial value problem
given f′(x) and f(a), find f(x) using definite integral
Displacement and Distance:
Displacement = ∫ab v(t) dt     Total distance = ∫ab |v(t)| dt

Total Change from a Rate:
∫ab f′(t) dt = f(b) − f(a)

Average Value:
f̄ = 1⁄(b−a) ∫ab f(x) dx

Recovering a function from its derivative:
f(x) = f(a) + ∫ax f′(t) dt
Worked Example 1: A particle moves with velocity v(t) = t² − 3t + 2 m/s for 0 ≤ t ≤ 3. Find (a) displacement and (b) total distance.

Find where v(t) = 0: t² − 3t + 2 = (t−1)(t−2) = 0 ⇒ t = 1, 2
Test signs: v(0.5) = 0.25 − 1.5 + 2 = 0.75 > 0; v(1.5) = 2.25 − 4.5 + 2 = −0.25 < 0; v(2.5) = 0.75 > 0

(a) Displacement = ∫03 (t²−3t+2) dt = [t³/3 − 3t²/2 + 2t]03
= (9 − 27/2 + 6) = 15 − 13.5 = 1.5 m (net movement to the right)

(b) Total distance: split at t = 1, 2
D1 = ∫01 (t²−3t+2) dt = [t³/3−3t²/2+2t]01 = 1/3−3/2+2 = 5/6
D2 = |∫12 (t²−3t+2) dt| = |[...]12| = |8/3−6+4−(1/3−3/2+2)| = |−1/6| = 1/6
D3 = ∫23 (t²−3t+2) dt = 5/6
Total distance = 5/6 + 1/6 + 5/6 = 11/6 ≈ 1.833 m
Worked Example 2: Water flows into a tank at the rate r(t) = 3 + 2t litres per minute. Find the total volume added from t = 0 to t = 5 minutes, and the average flow rate over this interval.

Total volume = ∫05 (3+2t) dt = [3t + t²]05 = 15 + 25 = 40 litres

Average flow rate = (1/5) ∫05 (3+2t) dt = 40/5 = 8 litres/min
Hot Tip: Displacement and total distance are different! Displacement is signed (can be negative) and equals ∫ v dt directly. Total distance is always positive — find where v changes sign, integrate each piece, and add the absolute values.

From Antiderivatives to Real-World Quantities

Integration is not just a technique for finding areas — it is the mathematical tool for recovering a quantity from its rate of change. This is one of the most powerful ideas in all of mathematics. Whenever we know how fast something is changing, integration lets us find how much it has changed.

The Fundamental Theorem of Calculus is the formal statement of this idea: if f′(t) is the rate of change of some quantity f, then ∫ab f′(t) dt = f(b) − f(a). In words: integrating a rate gives the total change.

Displacement vs. Total Distance: A Critical Distinction

When a particle moves along a line, its velocity v(t) can be positive (moving right/up) or negative (moving left/down). The integral of velocity gives displacement — the net change in position:

Displacement = ∫ab v(t) dt

But if the particle moves right for a while, then turns around and moves left, the distances in each direction partially cancel in the integral. Total distance counts all movement regardless of direction:

Total distance = ∫ab |v(t)| dt

To evaluate ∫|v(t)|dt, you must first find where v(t) = 0 (turning points), then split the integral at those values and add the absolute values of each piece.

Think of it physically: A person walks 3 metres east and then 2 metres west. Displacement = +1 m (net). Total distance = 5 m (actual walking). The displacement integral = 3 − 2 = 1; the distance integral = 3 + 2 = 5.

Total Change from a Rate of Change

Many real-world contexts give you a rate and ask for a total. This is straightforward: integrate the rate over the given interval.

Examples of this pattern:

  • Population growth: If dP/dt is the growth rate (people per year), then ∫010 dP/dt dt gives the total population change over 10 years.
  • Fluid flow: If r(t) is the flow rate (litres/second), then ∫0T r(t) dt gives the total volume that has flowed in T seconds.
  • Carbon emissions: If E(t) is the emission rate (tonnes/year), then ∫ab E(t) dt is the total carbon released from year a to year b.

The key insight is that the integral of a rate over time gives a total amount. The units confirm this: (litres/second) × seconds = litres.

Average Value of a Function

The average value of f(x) over [a, b] is defined as:

f̄ = 1⁄(b−a) ∫ab f(x) dx

Think of it this way: (b−a) is the width of the interval, and ∫ f(x) dx is the area. If you "spread" this area evenly over the interval, the height of the resulting rectangle is the average value.

This concept appears in physics (average velocity = displacement / time), economics (average cost), and engineering (average power). Note that average velocity = displacement/time = [∫ v dt] / (b−a), confirming the formula.

Initial Value Problems with Definite Integrals

Sometimes we are given f′(x) and an initial condition f(a) = c, and asked to find f(b) for some other value b. Using the FTC:

f(b) = f(a) + ∫ab f′(x) dx

Example: A factory's marginal cost is MC(x) = 0.02x + 5 dollars per unit. If the total cost of producing 0 units is $200 (fixed cost), find the total cost of producing 100 units.

C(100) = C(0) + ∫0100 (0.02x + 5) dx = 200 + [0.01x² + 5x]0100 = 200 + (100 + 500) = $800

Exam strategy: In rate-of-change problems, always state clearly what the integral represents before calculating. Write a sentence like "The total volume added equals ∫05 r(t) dt" before evaluating. This earns communication marks.

Connecting All the Ideas

These applications share a common structure: a quantity that varies continuously (velocity, flow rate, growth rate) is integrated to give a cumulative total. The definite integral always measures accumulated change. When you see phrases like "total amount", "how far", "how much has changed", or "average value" in a problem, integration is the tool.

For problems involving motion, set up a timeline: identify the time interval, find where velocity is zero, determine the sign of velocity in each sub-interval, and then integrate accordingly — summing absolute values for total distance, or integrating directly for displacement.

Common error: Treating displacement and total distance as the same. Always check whether v(t) changes sign on the interval. If it does, you must split the integral for total distance calculations.

Mastery Practice

  1. A particle moves with velocity v(t) = 4 − 2t m/s for 0 ≤ t ≤ 4. Find the displacement of the particle over this interval.
  2. Water flows into a tank at rate r(t) = 6 + t litres per hour. Find the total volume added from t = 0 to t = 4 hours.
  3. Find the average value of f(x) = x² + 1 on the interval [0, 3].
  4. A particle's velocity is v(t) = 3t² − 6t m/s. Find the total distance travelled from t = 0 to t = 3 seconds.
  5. A particle moves with velocity v(t) = t² − 4t + 3 m/s for 0 ≤ t ≤ 4. Find (a) the displacement and (b) the total distance travelled.
  6. The rate of change of a population of bacteria is dP/dt = 200e0.1t cells/hour, where t is in hours. If the initial population is 3000, find the population at t = 5 hours.
  7. The acceleration of a particle is a(t) = 6t − 2 m/s². If its initial velocity is v(0) = 1 m/s, find v(t) and the total distance travelled from t = 0 to t = 2 seconds.
  8. Find the average value of v(t) = sin t on [0, π]. Interpret what this means if v(t) is a velocity function.
  9. A car starts from rest and accelerates. Its velocity (m/s) at various times is recorded: v(0) = 0, v(2) = 8, v(4) = 14, v(6) = 18, v(8) = 20. Use the trapezoidal rule to estimate the total distance travelled over the first 8 seconds. Then find the average speed.
  10. The rate at which solar energy is absorbed by a roof panel is r(t) = 500 sin(πt/12) watts, where t is hours after 6 am (so t = 0 at 6 am, t = 12 at 6 pm). Find the total energy (in watt-hours) absorbed from 6 am to 6 pm, and the average power over this period.
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