When you push a book across a table, the work you do is simple: force times distance. But what if the force changes as you move? A spring gets harder to stretch the farther you pull it. Lifting a heavy cable becomes easier as more of it reaches the top. Pumping water from a tank requires more effort for the water at the bottom.
In these situations, the force isn't constant, so we can't just multiply. Instead, we break the motion into tiny pieces where the force is approximately constant, calculate the work for each piece, and add them up. This is exactly what a definite integral does.
The key insight: Work is the accumulation of force over distance, and integration is the mathematical tool for accumulation.
Quick self-check (30 seconds each). If you struggle with these, review the linked topics first.
1. Definite Integrals: Evaluate $\displaystyle\int_1^4 (2x + 3)\,dx$
$$\int_1^4 (2x + 3)\,dx = \left[x^2 + 3x\right]_1^4 = (16 + 12) - (1 + 3) = 24$$
✅ Got 24? Great—you're ready to proceed!
❌ Stuck? Review Definite Integrals before continuing.
2. Riemann Sums: What does the sum $\displaystyle\sum_{i=1}^{n} f(x_i^*)\Delta x$ approximate as $n \to \infty$?
It approximates the definite integral $\displaystyle\int_a^b f(x)\,dx$.
Geometrically, each term $f(x_i^*)\Delta x$ is the area of a thin rectangle. Adding them approximates the area under $y = f(x)$.
✅ Makes sense? You're ready!
❌ Fuzzy on this? Review Riemann Sums—the conceptual understanding is important for seeing why work becomes an integral.
| Property | Value |
|---|---|
| Section | Chapter 5, Section 4 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
When a constant force $F$ moves an object a distance $d$ in the direction of the force:
$$\boxed{W = Fd}$$
Units:
Important distinction: Weight is a force (already includes gravity), while mass must be multiplied by $g = 9.8 \text{ m/s}^2$ to get force.
Constant Force: Variable Force:
F f(x)
↓↓↓↓↓↓↓↓ ↓ ↓↓ ↓↓↓
──────── ────────
a b a b
W = F × (b-a) W = ∫ₐᵇ f(x) dxWhen force $f(x)$ varies with position as an object moves from $x = a$ to $x = b$:
$$\boxed{W = \int_a^b f(x) \, dx}$$
Divide the interval $[a, b]$ into $n$ small pieces of width $\Delta x$. On each piece, the force is approximately constant at $f(x_i^*)$, so the work on that piece is approximately:
$$W_i \approx f(x_i^*) \Delta x$$
Total work is the sum of all pieces:
$$W \approx \sum_{i=1}^{n} f(x_i^*) \Delta x$$
As $n \to \infty$ and $\Delta x \to 0$, this Riemann sum becomes the definite integral.
Still confused about Riemann sums? This is the same idea as approximating area under a curve with rectangles. Each rectangle has width $\Delta x$ and height $f(x_i^*)$. Review Riemann sums →
| Scenario | Force Function | Notes |
|---|---|---|
| Constant force | $f(x) = F$ | Lifting at constant speed |
| Linear force | $f(x) = kx$ | Springs (Hooke's Law) |
| Inverse square | $f(x) = \frac{k}{x^2}$ | Gravity at varying distances |
| Position-dependent | $f(x) = g(x)$ | Given force function |
| Mistake | Why It Happens | How to Avoid |
|---|---|---|
| Confusing mass and weight | Given "15 kg" and using 15 as the force | Mass (kg) needs $F = mg$. Weight (lb, N) is already a force. |
| Wrong limits of integration | Mixing up starting and ending positions | Draw a diagram. Label positions clearly before setting up the integral. |
| Forgetting units | Getting a number without checking reasonableness | Always write units. Work = force × distance, so J = N·m or ft-lb. |
| Using $W = Fd$ for variable force | Force changes but you multiply anyway | If $f$ depends on $x$, you must integrate: $W = \int f(x)\,dx$ |
The concept of work was formalized in the early 1800s by French mathematician Gaspard-Gustave Coriolis (yes, the Coriolis effect is named after him). He defined work as "weight lifted through a height" to analyze the efficiency of machines and waterwheels. The integral formulation came from engineers trying to calculate how much useful output they could get from steam engines—the more work per unit of fuel, the better the engine.
A worker lifts a 15-kg toolbox from the ground to a shelf 1.8 m high. How much work is done? Use $g = 9.8 \text{ m/s}^2$.
A particle moves along the $x$-axis from $x = 0$ to $x = 4$ meters. A force of $f(x) = 3x + 2$ newtons acts on it. Find the work done.
A repelling magnetic force of $f(x) = \frac{12}{x^2}$ newtons acts on a charged particle, where $x$ is the distance in meters from a fixed magnet. How much work is done in moving the particle from $x = 2$ m to $x = 6$ m?
A 50-foot chain weighing 3 lb/ft hangs from the top of a building. How much work is required to pull the entire chain to the top?
A toy slingshot exerts a restoring force $f(x) = 80x$ N when the pouch is pulled back $x$ meters from equilibrium. A 50-gram projectile (mass = 0.05 kg) is placed in the pouch, pulled back 0.12 m, and released from rest.
The graph shows force $f(x)$ (in newtons) versus position $x$ (in meters) for an object moving from $x = 0$ to $x = 5$.
f(x)
8 | ___________
| /
4 |___/
|
0 +---+---+---+---+---→ x
0 1 2 3 4 5Which statement best describes the work done?
Work as Accumulated Effort:
Think of work like filling a bucket with water one cup at a time. Each cup is a small bit of effort (force × tiny distance). The total work is adding up all those small efforts. When force varies, some cups are heavier than others, but we still add them all up. The integral is just a precise way of doing this addition when there are infinitely many infinitely small cups.
Mount Everest is 8,849 m tall. For a 70 kg climber, the work against gravity alone is: $$W = mgh = (70)(9.8)(8849) \approx 6.1 \times 10^6 \text{ J} = 6.1 \text{ MJ}$$
That's equivalent to about 1,450 food Calories—roughly a day's worth of eating! (In reality, climbers burn far more because of inefficiency, cold, and carrying gear.)
Looking back:
Looking ahead:
Real-world connections:
| Previous | Up | Next |
|---|---|---|
| Section Overview | Ch5 Sec4 Skills | Spring Work |
Last updated: 2026-01-23