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M11.4MechanicsStretch

Using Integration

Integration reverses differentiation, so it steps back up the kinematics chain: integrate acceleration to get velocity, and velocity to get displacement. The constant of integration is found from initial conditions.

25 min Variable Acceleration
Edexcel Mechanics Y1 — Variable Acceleration playlist (Zeeshan Zamurred)Curated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Integrate acceleration to find velocity
  • Integrate velocity to find displacement
  • Use initial conditions to find the constant
  • Find distance travelled with definite integrals
1

The integration chain

Integration goes the opposite way to differentiation: integrate aa to get vv, and integrate vv to get ss. Each integration introduces a constant, found from a known value (e.g. initial velocity).

v=∫a dt,s=∫v dtv = \int a \, dt, \qquad s = \int v \, dt
Integrate up: a→v→sa \to v \to s (with +c+c).
Example 1Acceleration a=6ta = 6t, initial velocity 4 m/s. Find vv.
1v=∫6t dt=3t2+cv = \int 6t \, dt = 3t^2 + c.
2At t=0t = 0, v=4v = 4, so c=4c = 4.
Answerv=3t2+4v = 3t^2 + 4
2

Using initial conditions

Each +c+c is found by substituting a known value — usually the value at t=0t = 0. Without this step the answer is incomplete.

Tip — Every integration needs its + c pinned down by an initial condition.

3

Distance with definite integrals

The distance travelled between two times is the definite integral of velocity: ∫t1t2v dt\int_{t_1}^{t_2} v \, dt. (If the velocity changes sign, split the integral at v=0v = 0 and add the magnitudes, as with areas under curves.)

s=∫t1t2v dts = \int_{t_1}^{t_2} v \, dt
Definite integral gives displacement over the interval.

Formula recap

v=∫a dtv = \int a \, dt
Velocity from acceleration.
s=∫v dts = \int v \, dt
Displacement from velocity.
s=∫t1t2v dts = \int_{t_1}^{t_2} v \, dt
Displacement over an interval.

Common mistakes to avoid

Forgetting the constant of integration.
Find + c from the initial conditions every time.
Integrating to go from velocity to acceleration.
Integration goes a → v → s; differentiation goes the other way.

Key takeaways

  • Integrate to step up: a → v → s.
  • Use initial conditions to find each constant of integration.
  • Distance over an interval = definite integral of velocity.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Check your understanding

Work through questions on Using Integration. Use the feedback to decide what needs another look.

1 past-paper question · 2 marks

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