Chapter 14

Applications of Integration and Kinematics

Using integration to find areas under curves and to relate displacement, velocity and acceleration in kinematics.

Area under a curve

A definite integral gives the area between a curve and the x-axis between two limits. The area under y = f(x) from x = a to x = b is ∫ from a to b of f(x) dx.

Key idea

Area = ∫ab f(x) dx. In kinematics, differentiation and integration link displacement s, velocity v and acceleration a: v = ds⁄dt, a = dv⁄dt, and reversing, v = ∫ a dt, s = ∫ v dt.

Kinematics

Given the acceleration, integrate to obtain velocity; given the velocity, integrate to obtain displacement. Use given conditions (such as v = 0 at t = 0) to find each constant. A particle is at rest when v = 0.

Worked example

A particle has velocity v = 4t. The displacement from t = 0 to t = 3 is ∫ from 0 to 3 of 4t dt = [2t2] from 0 to 3 = 18 − 0 = 18 m.

Remember

  • Differentiate s → v → a; integrate a → v → s.
  • The particle is at rest when v = 0.
  • Use initial conditions to find constants of integration.

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