Tags: work integral, variable force, Hooke's law, spring constant, pumping water, lifting cable, average value of a function, mean value theorem for integrals, Texas A&M calculus
Work is force times distance, but when force varies with position, you integrate: W = ∫ F(x) dx. The three main problem types are spring problems (Hooke's law), cable/rope lifting, and pumping water out of a tank. The average value of a function on an interval is the integral divided by the interval length, and the Mean Value Theorem for integrals guarantees the function hits its average at some point.
Work
The energy transferred by a force acting over a distance. When force is constant: W = F · D. When force varies: W = ∫ F(x) dx.
Hooke's law
The restoring force of a spring is proportional to displacement from its natural length: F(x) = −k(x − x₀). The magnitude of force needed to hold the spring at extension x is k(x − x₀). Valid for small displacements.
Spring constant (k)
The proportionality constant in Hooke's law. Measured in N/m or lb/ft. A stiffer spring has a larger k.
Linear weight density (δ)
Weight per unit length of a rope or cable (e.g. lb/ft). Used when computing work to lift a hanging rope.
Weight density (ρ)
Weight per unit volume of a fluid. For water: ρ = 1000 kg/m³ (with g = 9.8 m/s²) in metric, or 62.5 lb/ft³ in imperial.
Average value of a function
f_ave = [1/(b − a)] · ∫ from a to b of f(x) dx. It is the constant height of a rectangle with the same base and same area as the region under the curve.
Mean Value Theorem for integrals
If f is continuous on [a, b], there exists some c in [a, b] such that ∫ from a to b of f(x) dx = f(c)(b − a). Equivalently, f(c) = f_ave for at least one point c.
When force depends on position, slice the motion into tiny intervals dx, where force is approximately constant.
The work done over each slice is F(x) dx.
Total work:
W = ∫ from a to b of F(x) dx
Units: if force is in Newtons and distance in metres, work is in Joules (N·m). If force is in pounds and distance in feet, work is in foot-pounds (ft·lb).
A spring at rest has natural length x₀. Stretching or compressing it to length x requires a force of magnitude k|x − x₀|.
To find k: use a known force-displacement pair. If a 24 N force stretches a spring from 1 m to 1.8 m, then 24 = k(0.8), so k = 30 N/m.
To find work to stretch from one length to another, let u = displacement from natural length. Then:
W = ∫ from u₁ to u₂ of k · u · du
For the example above, stretching from 1.5 m to 2 m means u goes from 0.5 to 1.0:
W = ∫ from 0.5 to 1.0 of 30u du = 30[u²/2] from 0.5 to 1.0 = 15(1 − 0.25) = 11.25 J
A cable of total weight W_total and length L hangs vertically. Linear weight density δ = W_total / L.
A small piece of rope at height h from the bottom must be lifted a distance D(h) to reach the top.
Total work:
W = ∫ from 0 to L of D(h) · δ · dh
For pulling the entire cable to the top of a building: the piece at height h (measured from top) must be lifted h units. So W = ∫₀^L δ · h · dh = δ · L²/2.
For pulling up only half: the piece at distance h below the top (for h from 0 to L/2) is lifted h. The remaining half stays put. W = ∫₀^(L/2) δ · h · dh.
Slice the water into thin horizontal layers at height h.
Each layer has volume A(h) · dh, where A(h) is the cross-sectional area at height h.
Weight of the layer: ρ · A(h) · dh.
Distance to pump this layer to the top (or to the top of a spout): D(h).
Total work:
W = ∫ from bottom to top of D(h) · ρ · A(h) · dh
The cross-sectional area A(h) depends on the tank shape. For a rectangular tank it is constant. For a cone or sphere it varies with h.
Rectangular tank (10 m × 3 m × 2 m): A(h) = 10 × 3 = 30 m² at every height. If pumping from the top, D(h) = 2 − h (where h is measured from the bottom). W = ∫₀² ρg(2 − h)(30) dh.
With a spout of height s above the tank top, D(h) = (tank height + s) − h.
Inverted cone (height 8 m, top radius 5 m): at height h from the bottom, the radius is r(h) = 5h/8 (by similar triangles), so A(h) = π(5h/8)².
Half-cylinder trough (length 6 ft, radius 2 ft): requires setting up the cross-section width from the circular equation.
To pump out the top half of the water from a full tank, integrate from the midpoint to the top.
To pump out the bottom half (starting from a full tank), integrate from the bottom to the midpoint.
These give different amounts of work because the bottom water must be lifted further.
The average value of f on [a, b] is:
f_ave = [1/(b − a)] · ∫ from a to b of f(x) dx
Rearranging: ∫ from a to b of f(x) dx = (b − a) · f_ave. The integral equals the area of a rectangle with width (b − a) and height f_ave.
Example: the average value of 1/x on [1, 4] is (1/3) · ∫₁⁴ (1/x) dx = (1/3) ln 4.
A car with variable velocity v(t) from t = a to t = b has average velocity:
v_ave = [1/(b − a)] · ∫ from a to b of v(t) dt
This equals displacement divided by time: [s(b) − s(a)] / (b − a), since s'(t) = v(t).
If f is continuous on [a, b], there is at least one c in [a, b] where f(c) equals the average value.
Intuition: if your average speed during a drive is 73 mph, at some moment your speed was exactly 73 mph.
This follows from the ordinary Mean Value Theorem applied to an antiderivative F of f: F(b) − F(a) = F'(c)(b − a) = f(c)(b − a).
Work (variable force):
W = ∫ from a to b of F(x) dx
Hooke's law (spring force):
F = k · (displacement from natural length)
Work on a spring:
W = ∫ from u₁ to u₂ of k · u · du = k(u₂² − u₁²)/2
Lifting a cable:
W = ∫ from 0 to L of D(h) · δ · dh
Pumping water:
W = ∫ from bottom to top of D(h) · ρ · A(h) · dh
Average value:
f_ave = [1/(b − a)] · ∫ from a to b of f(x) dx
Mean Value Theorem for integrals:
∫ from a to b of f(x) dx = f(c)(b − a) for some c in [a, b]
⚠️ In spring problems, measure displacement from the natural length, not from zero. If natural length is 1 m and you stretch to 2 m, the displacement is 1 m, not 2 m.
⚠️ In pumping problems, set up your coordinate system clearly and write D(h) carefully. A common error is getting the distance function backwards (measuring from the wrong end of the tank).
⚠️ A spout adds to the pumping distance. If there is a 0.5 m spout, every layer of water must be lifted an additional 0.5 m.
⚠️ Pumping out the top half of a tank requires less work than pumping out the bottom half, because the bottom water sits further from the exit.
⚠️ The Mean Value Theorem for integrals requires continuity. It is the integral version of the ordinary MVT and is a standard exam proof or short-answer question.
⚠️ Watch the units. Metric water-pumping problems use ρ = 1000 kg/m³ and need a factor of g = 9.8 m/s² to convert mass density to weight (force). Imperial problems use 62.5 lb/ft³ directly as weight density.
Q: If f(x) = 2x² + 1 (in Newtons), find the work to move an object from x = 1 m to x = 2 m.
A: W = ∫₁² (2x² + 1) dx = [2x³/3 + x] from 1 to 2 = (16/3 + 2) − (2/3 + 1) = 14/3 + 1 = 17/3 J, which is approximately 5.67 J.
Q: A spring has natural length 1 m and k = 30 N/m. How much work to stretch it from 1.5 m to 2 m?
A: Displacement goes from 0.5 to 1.0 m. W = ∫₀.₅¹ 30u du = 30[u²/2] from 0.5 to 1 = 15(1 − 0.25) = 11.25 J.
Q: A 200 lb cable, 100 ft long, hangs vertically. How much work to pull the entire cable to the top?
A: δ = 200/100 = 2 lb/ft. Let h = distance from the top (0 to 100). The piece at distance h must be lifted h ft. W = ∫₀¹⁰⁰ 2h dh = 2[h²/2]₀¹⁰⁰ = 10,000 ft·lb.
Q: A rectangular tank (10 × 3 × 2 m) full of water has a 0.5 m spout. How much work to empty a quarter of the tank from the top?
A: A quarter of 2 m depth is 0.5 m, so pump water from h = 1.5 to h = 2 (top). The exit is at height 2.5 m (tank top + spout). D(h) = 2.5 − h. W = ∫₁.₅² (9800)(30)(2.5 − h) dh.
Q: State the Mean Value Theorem for integrals and give a real-world interpretation.
A: If f is continuous on [a, b], then f(c) = [1/(b − a)] ∫ₐᵇ f(x) dx for some c in [a, b]. In words: a continuous function must equal its average value at some point. For a car trip, at some instant your speed equals your average speed for the journey.
Q: What is the average value of 1/x on [1, 4]?
A: f_ave = (1/3) ∫₁⁴ (1/x) dx = (1/3)(ln 4 − ln 1) = (ln 4)/3 ≈ 0.462.
work integral, variable force, Hooke's law, spring constant, spring work, pumping water, tank problem, lifting cable, weight density, linear density, foot-pounds, joules, Newton-metres, average value, mean value theorem for integrals, MVT integrals, average velocity, continuous function, MATH 151, MATH 152, Texas A&M