Chapter Contents

• Applications of Integration
• 1. Applications of the Indefinite Integral
• 2. Area Under a Curve
• 3. Area Between 2 Curves
• 4. Volume of Solid of Revolution
• 5. Centroid of an Area
• 6. Moments of Inertia
• 7. Work by a Variable Force
• 8. Electric Charges
• 9. Average Value
• Head Injury Criterion
• a. Normal Braking
• b. Crash Tests
• c. Data
• d. Severity Index
• e. Head Injury Criterion
• f. Modelling the HIC
• g. Conclusions
• 10. Force by Liquid Pressure
• 11. Arc Length of a Curve
• 12. Arc Length of a Curve in Parametric or Polar Coordinates
• Applications of Integration Problem Solver

• Comments, Questions?

Feelings about Math

Recommendation

Easy to understand calculus lessons on DVD. Try before you commit. More info:
MathTutorDVD.com

Online Algebra Solver

Solve your algebra problem step by step!

Online Algebra Solver »

Share

Share this page.

From the math blog...

1. Applications of the Indefinite Integral

by M. Bourne

Displacement from Velocity, and Velocity from Acceleration

A very useful application of calculus is displacement, velocity and acceleration.

Recall (from Derivative as an Instantaneous Rate of Change) that we can find an expression for velocity by differentiating the expression for displacement:

Similarly, we can find the expression for the acceleration by differentiating the expression for velocity, and this is equivalent to finding the second derivative of the displacement:

It follows (since integration is the opposite process to differentiation) that to obtain the displacement, s of an object at time t (given the expression for velocity, v) we would use:

Similarly, the velocity of an object at time t, given the acceleration a, is given by:

Example 1

A proton moves in an electric field such that its acceleration (in cms^-2) is

a = -20(1+2t)^-2, where t is in seconds.

Find the velocity as a function of time if v = 30 cms^-1 when t = 0.

Answer

Example 2

A flare is ejected vertically upwards from the ground at 15 m/s. Find the height of the flare after 2.5 s.

Answer

Displacement and Velocity Formulas

Using integration, we can obtain the well-known expressions for displacement and velocity, given a constant acceleration a, initial displacement zero, and an initial velocity v₀:

v = ∫ a dt

v = at + K

Since the velocity at t = 0 is v₀. we have K = v₀. So:

v = v₀ + at

Similarly,

s = ∫v dt = ∫(v₀ + at) dt

s = v₀t + at²/2 + C

Since the displacement at t = 0 is s = 0, we have C = 0. So:

Voltage across a Capacitor

Definition: The current, i (amperes), in an electric circuit equals the time rate of change of the charge q, (in coulombs) that passes a given point in the circuit. We can write this (with t in seconds) as:

By writing i dt = dq and integrating, we have:

The voltage, V_C (in volts) across a capacitor with capacitance C (in farads) is given by

.

It follows that

You can see some more advanced applications of this at Applications of Ordinary Differential Equations.

Example 3

The electric current (in mA) in a computer circuit as a function of time is i = 0.3 − 0.2t. What total charge passes a point in the circuit in 0.050s?

Answer

Example 4

The voltage across an 8.50 nF capacitor in an FM receiver circuit is zero. Find the voltage after 2.00 μs if a current i = 0.042t (in mA) charges the capacitor.

Answer