Work Calculus Calculator — W = ∫F(x)dx
Calculate work done by variable forces using the definite integral W=∫F(x)dx — solving spring work problems (Hooke's law), chain/rope lifting problems, and liquid pumping problems that are standard in Calculus 2 courses — with full symbolic integration steps using the Fundamental Theorem of Calculus.
Work Done by Variable Force — W = ∫F(x)dx
Shaded area under the F(x) curve from x=a to x=b equals the work done W = ∫F(x)dx
x = 0 is natural length · positive x = stretched · negative x = compressed
Spring Work — W = ½k(x₁² − x₀²)
Chain Lifting Work — W = ∫ρg(L−x)dx
Pumping Work — W = ∫ρg·A(x)·(H−x)dx
Enter any F(x) function and bounds to visualize the area under the force-distance curve — this area IS the work done. The shaded region equals W = ∫F(x)dx — the work integral.
Work = Area Under Force-Distance Graph
Violet shading = positive work (force and displacement in same direction) · Red = negative work
Work as an Integral — Why W = ∫F(x)dx
This work calculus calculator computes work done by variable forces using the work integral W=∫F(x)dx — the fundamental connection between force and energy in calculus. When force is constant, work is simply W = F × d. But when force varies with position — as in spring compression, chain lifting, and liquid pumping — you need the integral of force over the displacement to get the exact work done.
The idea is elegantly simple: divide the path from x=a to x=b into tiny segments dx. On each segment, the force is approximately constant at F(x). The work done on that tiny segment is dW = F(x)dx. Total work = integral of force over all segments:
This is the definition of work done by a variable force: W=∫F(x)dx is not just a formula — it is the mathematical limit of summing infinitely many infinitesimal work contributions. The work equation integral answers the question: is work the integral of force? Yes — work is the integral of force with respect to distance.
Geometric meaning: work = area under the F(x) curve from x=a to x=b. When force and displacement point in the same direction (F(x) > 0), the area is positive and the work done is positive. When they oppose each other (F(x) < 0), the area is negative and work done is negative. The work formula calculus makes this precise through the definite integral.
The Force-Distance Graph — Area = Work
The F vs x graph (force-distance graph) is the central visual for understanding variable force work. Every shape under this graph tells a physical story:
- Constant force (horizontal line): Rectangle with area F × d = W. The simplest case: W = Fd.
- Spring force F=kx (line through origin): Right triangle with area ½ × base × height = ½ × d × (kd) = ½kd². This is exactly the spring work formula.
- Variable force (any curve): Irregular area under the curve — computed by the work integral W = ∫F(x)dx.
Positive area (F and displacement same direction) = positive work done on the object. Negative area = negative work (force opposes motion). The area under the force-distance graph is always equal to the total work done by that force — this is why work calculus is so powerful.
Is work the integral of force? Yes. W = ∫[a to b] F(x) dx is the work done by a variable force F(x) as an object moves from position x=a to x=b. The integral of force with respect to distance gives work in Joules (N·m = J).
Spring Work — Hooke's Law W = ½kd²
The spring work problem is the most important example in work calculus 2. Hooke's Law states: F(x) = k × x — the spring force is proportional to displacement from natural length, where k is the spring constant (N/m). At x=0, force is zero; at x=d, force is kd. The force increases linearly.
Why the factor ½? Because force starts at 0 and ends at kd — the average force is kd/2. Work = average force × distance = (kd/2) × d = ½kd². This connects perfectly to the triangular area under the F=kx line. For non-zero starting positions: W = ½k(x₁² − x₀²).
Example 1 — Spring Work: k=150 N/m, compress from 0 to 0.2m
- Hooke's Law: F(x) = 150x
- Work integral: W = ∫[0 to 0.2] 150x dx
- Antiderivative: ∫150x dx = 75x²
- Evaluate: W = [75x²]₀^0.2 = 75(0.04) − 0 = 3 J
- Check: W = ½ × 150 × (0.2)² = ½ × 150 × 0.04 = 3 J ✓
Example 2 — Spring Work: k=300 N/m, stretch from 0.1m to 0.4m
- F(x) = 300x
- W = ∫[0.1 to 0.4] 300x dx = [150x²]₀.₁^0.4
- W = 150(0.16) − 150(0.01) = 24 − 1.5 = 22.5 J
- Check: W = ½ × 300 × (0.16 − 0.01) = 150 × 0.15 = 22.5 J ✓
Example 3 — Car Suspension: k=1000 N/m, compress 0 to 0.05m
- F(x) = 1000x, W = ∫[0 to 0.05] 1000x dx
- W = [500x²]₀^0.05 = 500(0.0025) = 1.25 J
Lifting a Chain — Linear Mass Density Problems
The lifting chain problem is a classic calculus 2 work problem. A chain of length L and linear mass density ρ (kg/m) lies on the ground. As you lift it, the force needed decreases — less chain remains to be lifted. This is why work done by variable force requires the work integral.
Let x = distance the bottom of the chain has been lifted. The remaining hanging length is (L − x), with weight F(x) = ρg(L − x). The work integral for lifting the entire chain:
Example 1 — Chain: L=10m, ρ=3 kg/m, lift entire chain
- F(x) = ρg(L−x) = 3 × 9.807 × (10−x) = 29.42(10−x) N
- W = ∫[0 to 10] 29.42(10−x) dx = 29.42[10x − x²/2]₀^10
- W = 29.42(100 − 50) = 29.42 × 50 = 1471 J
- Check: W = ½ρgL² = ½ × 3 × 9.807 × 100 = 1471 J ✓
- Alternative: lift center of mass (at L/2 = 5m): W = (30kg)(9.807)(5m) = 1471 J ✓
Example 2 — Chain: L=10m, ρ=3 kg/m, lift bottom 4m only
- d = 4m (partial lift), F(x) = 29.42(10−x)
- W = ∫[0 to 4] 29.42(10−x) dx = 29.42[10x − x²/2]₀^4
- W = 29.42(40 − 8) = 29.42 × 32 = 941.4 J
Example 3 — Chain: L=10m, ρ=2 kg/m, lift entire chain
- W = ½ρgL² = ½ × 2 × 9.807 × 100 = 980.7 J
- Equivalent to lifting center of mass: W = (20kg)(9.807)(5m) = 980.7 J ✓
Pumping Liquid — Tank Geometry Problems
The pumping liquid problem is the hardest standard work calculus 2 problem because it requires deriving F(x) from the tank geometry. The key formula is:
Example 1 — Cylindrical Tank: r=2m, water to h=5m, pump to H=7m
- A(x) = πr² = π(4) = 4π m² (constant)
- W = ρg·A·∫[0 to 5] (7−x) dx = 1000×9.807×4π×[7x−x²/2]₀^5
- = 123,271 × (35 − 12.5) = 123,271 × 22.5 = 2,773,597 J ≈ 2.77 MJ
Example 2 — Conical Tank: apex at bottom, R=4m at H=6m, full of water, pump to 6m
- At height x: radius r(x) = R·x/H_tank = 4x/6 = 2x/3
- A(x) = π(2x/3)² = 4πx²/9
- W = ρg·(4π/9)·∫[0 to 6] x²(6−x) dx
- ∫x²(6−x) dx = ∫(6x²−x³) dx = 2x³ − x⁴/4
- Evaluated 0 to 6: 2(216) − 1296/4 = 432 − 324 = 108
- W = 1000×9.807×(4π/9)×108 ≈ 1,478,000 J ≈ 1.48 MJ
Example 3 — Rectangular Tank: 3m×4m, water to h=5m, pump to H=6m
- A(x) = 3 × 4 = 12 m² (constant)
- W = 1000×9.807×12×∫[0 to 5] (6−x) dx = 117,684×[6x−x²/2]₀^5
- = 117,684×(30 − 12.5) = 117,684×17.5 = 2,059,470 J ≈ 2.06 MJ
How to Set Up a Work Integral — Step-by-Step
Follow this five-step method for any work calculus problem — spring, chain, pumping, or variable force:
- Draw a diagram and label coordinates. Choose x as the direction of motion. Label the start and end positions.
- Identify F(x) — the force as a function of position. For springs: F(x)=kx. For chains: F(x)=ρg(L−x). For pumping: F(x)=ρg·A(x)·(H−x).
- Identify limits of integration a and b — where motion starts and ends.
- Write the work integral: W = ∫[a to b] F(x) dx
- Evaluate using the Fundamental Theorem of Calculus: find the antiderivative, evaluate at b, subtract value at a.
| Problem Type | F(x) expression | Limits | Work integral W=∫F(x)dx |
|---|---|---|---|
| Spring (Hooke's Law) | kx | [x₀, x₁] | ½k(x₁²−x₀²) |
| Chain lifting (full) | ρg(L−x) | [0, L] | ½ρgL² |
| Chain lifting (partial) | ρg(L−x) | [0, d] | ρg(Ld−d²/2) |
| Pump (cylinder) | ρg·πr²·(H−x) | [0, h] | ρgπr²(Hh−h²/2) |
| Pump (cone, apex↓) | ρg·π(Rx/H_t)²·(H−x) | [0, h] | ρgπ(R/H_t)²·(Hh³/3−h⁴/4) |
| Variable force | any F(x) | [a, b] | ∫[a to b] F(x) dx |
Worked Examples — 8 Complete Calculus 2 Work Problems
1. Spring: k=150 N/m, compress from 0 to 0.2m
- F(x) = 150x, W = ∫[0 to 0.2] 150x dx = [75x²]₀^0.2
- W = 75(0.04) − 0 = 3 J
2. Spring: k=300 N/m, stretch from 0.1m to 0.4m
- W = ∫[0.1 to 0.4] 300x dx = [150x²]₀.₁^0.4 = 150(0.16)−150(0.01)
- W = 24 − 1.5 = 22.5 J
3. Chain: L=10m, ρ=3 kg/m, lift entirely
- W = ½ρgL² = ½ × 3 × 9.807 × 100 = 1471 J
4. Chain: L=10m, ρ=3 kg/m, lift bottom 4m
- W = ρg(Ld − d²/2) = 3×9.807×(10×4 − 16/2) = 29.42×32 = 941.4 J
5. Cylindrical tank: r=2m, h=5m water, pump to H=7m above bottom
- A=4π, W = 1000×9.807×4π×(7×5 − 25/2) = 123,271×22.5 = 2,773,597 J ≈ 2.77 MJ
6. Conical tank: apex at bottom, base R=4m at height 6m, full, pump to 6m
- A(x) = π(4x/6)² = 4πx²/9
- W = 1000×9.807×(4π/9)×∫[0 to 6] x²(6−x)dx = 9807×4π/9×108 ≈ 1,478,000 J
7. Variable force: F(x)=3x²+2x from x=1 to x=3
- W = ∫[1 to 3] (3x²+2x) dx = [x³+x²]₁^3
- W = (27+9) − (1+1) = 36 − 2 = 34 J
8. Variable force: F(x)=10sin(πx/2) from x=0 to x=1
- W = ∫[0 to 1] 10sin(πx/2) dx = 10×[−cos(πx/2)/(π/2)]₀^1
- W = (20/π)[−cos(π/2)+cos(0)] = (20/π)[0+1] = 20/π ≈ 6.366 J
Common Mistakes in Calculus Work Problems
Mistake 1 — Wrong coordinate system for pumping problems
- ❌ Wrong: x measured inconsistently — sometimes from top, sometimes from bottom
- ✅ Correct: Always define x explicitly (e.g., "x = height from tank bottom, 0 ≤ x ≤ h")
Mistake 2 — Wrong "distance to pump" expression
- ❌ Wrong: If x is height from bottom and pump is at height H, using distance = x
- ✅ Correct: Distance to pump = H − x (liquid at height x must travel H − x upward)
Mistake 3 — Chain weight not decreasing as chain is lifted
- ❌ Wrong: Using F = ρgL (constant) for the entire lift
- ✅ Correct: F(x) = ρg(L−x) — as x increases, less chain hangs below, force decreases
Mistake 4 — Wrong limits of integration
- ❌ Wrong: ∫[0 to H] for pumping when tank is only filled to height h < H
- ✅ Correct: ∫[0 to h] — limits represent the physical range of the liquid, not the pump height
Mistake 5 — Forgetting ρ×g (using mass instead of weight)
- ❌ Wrong: W = ∫ρ·A(x)·(H−x) dx (missing g)
- ✅ Correct: W = ∫ρg·A(x)·(H−x) dx — weight = mass × g = ρ×A×dx×g