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Mom's Basement University

Work and Average Value

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn two practical applications of definite integrals: calculating the work done by a variable force and finding the average value of a function over an interval.

The Concept

Section titled “The Concept”

Work

Section titled “Work”

When a force varies with position, the work done moving an object from x = a to x = b is

W=∫abF(x) dxW = \int_a^b F(x)\,dx

If the force is constant, this just gives W = F times d (force times distance). But when the force changes, like with a spring or gravity at different distances, you need the integral.

A common example is Hooke’s Law: the force to stretch a spring is F(x) = kx, where k is the spring constant and x is the displacement from the natural length.

Average Value

Section titled “Average Value”

The average value of a continuous function f(x) on [a, b] is

favg=1b−a∫abf(x) dxf_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx

Think of it this way: the integral gives the total area under the curve. Dividing by the width of the interval gives the height of a rectangle with the same area. That height is the average value.

Worked Examples

Section titled “Worked Examples”

Example 1: Work done by a spring

A spring requires 10 N to stretch it 0.1 m beyond its natural length. How much work is done stretching it 0.5 m?

First find k from Hooke’s Law: F = kx, so 10 = k(0.1), giving k = 100 N/m.

W=∫00.5100x dx=100[x22]00.5=50(0.25)=12.5 JW = \int_0^{0.5} 100x\,dx = 100\left[\frac{x^2}{2}\right]_0^{0.5} = 50(0.25) = 12.5 \text{ J}

Example 2: Average value of a function

Find the average value of f(x) = x² on the interval [0, 3].

favg=13−0∫03x2 dx=13[x33]03=13⋅273=93=3f_{\text{avg}} = \frac{1}{3 - 0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3}\cdot\frac{27}{3} = \frac{9}{3} = 3

So the average value is 3. That means a rectangle with base 3 and height 3 has the same area (9) as the region under the parabola from 0 to 3.

Example 3: Work done by a variable force

A force F(x) = 4x - x² (in newtons) acts on an object as it moves from x = 1 to x = 4 meters. Find the total work done.

W=∫14(4x−x2) dx=[2x2−x33]14W = \int_1^4 (4x - x^2)\,dx = \left[2x^2 - \frac{x^3}{3}\right]_1^4

Evaluate at the bounds:

(2(16)−643)−(2(1)−13)=(32−643)−(2−13)\left(2(16) - \frac{64}{3}\right) - \left(2(1) - \frac{1}{3}\right) = \left(32 - \frac{64}{3}\right) - \left(2 - \frac{1}{3}\right) =96−643−6−13=323−53=273=9 J= \frac{96 - 64}{3} - \frac{6 - 1}{3} = \frac{32}{3} - \frac{5}{3} = \frac{27}{3} = 9 \text{ J}

The force starts at F(1) = 3 N, peaks at F(2) = 4 N, and drops to F(4) = 0 N. The integral captures the total accumulated work across that changing force.

Real-World Application

Section titled “Real-World Application”

Work and average value show up constantly in physics and engineering:

  • Springs, pumps, and lifting problems all involve variable forces where you need the integral to get total work
  • Average value is used in signal processing to find the mean of a continuous signal over time
  • Game physics engines calculate work done by forces to determine energy changes in objects
  • Temperature modeling uses average value to find the mean temperature over a day or season

Quiz

Section titled “Quiz”
The work done by a variable force F(x) from a to b is
A spring has constant k = 200 N/m. The work to stretch it 0.3 m from its natural length is
The average value formula divides the integral by
The average value of f(x) = 6x on [0, 4] is
If the work to stretch a spring 0.5 m is 12.5 J, the spring constant k is