Section 6.5 closes the "applications of the integral" thread from Weeks 9-10 with work and fluid force. Sections 7.1-7.3 then open a new thread: functions defined by inverting other functions, which is exactly how the natural logarithm and exponential function get their rigorous definitions in this course.
11.1 Work Done by a Variable Force
For a constant force F moving an object a distance d, work is simply W=Fd. When the force F(x) varies with position, the same "approximate with thin pieces, then integrate" strategy from Weeks 7 and 10 applies: over a short displacement dx, the force is nearly constant, contributing work F(x)dx. Summing and taking the limit:
Example 11.1 — Work to stretch a spring
By Hooke's Law, the force needed to stretch a spring x metres past its natural length is F(x)=kx. If k=200 N/m, find the work to stretch it from 0 to 0.5 m.
Step 1 — Apply Eq. 11.1
W = ∫00.5 200x dx
Step 2 — Evaluate
200[x2/2]00.5 = 100(0.25)
W = 25 J
11.2 Work Done Against Gravity (Pumping and Lifting)
To pump liquid out of a tank, slice the liquid into thin horizontal layers. Each layer at depth x has weight (density × gravity × cross-sectional area × dx) and must be lifted a distance depending on x; the total work is the integral of (weight of layer) × (distance lifted) over the tank's depth.
Example 11.2 — Pumping water from a tank
A cylindrical tank of radius 2 m and height 5 m, full of water (density 1000 kg/m³, g=9.8 m/s²), is pumped out over the top. Find the work done.
Step 1 — Set up a layer at depth x from the top
A layer at depth x (0≤x≤5) has volume π(2)2dx = 4π dx, weight 1000(9.8)(4π)dx, and must be lifted a distance x to reach the top.
Step 2 — Integrate weight × distance
W = ∫05 39200π x dx = 39200π[x2/2]05
W = 39200π(12.5) = 490000π J ≈ 1.54×106 J
11.3 Fluid Force
The force exerted by a fluid on a submerged vertical surface follows the same slicing idea: at depth x, pressure is (density)(g)(x), and the force on a thin horizontal strip of width L(x) and thickness dx is pressure × area:
— the same "slice, multiply, integrate" pattern as work, applied to force instead.
11.4 Inverse Functions and the Horizontal Line Test
A function f has an inverse f−1 exactly when f is one-to-one: no two different inputs share an output. Graphically, this is the horizontal line test — the mirror image of the vertical line test from Week 1 §1.2. The inverse undoes f:
and its graph is the reflection of f's graph across the line y=x — swapping the roles of x and y is exactly what "inverse" means.
A function that is increasing throughout its domain (or decreasing throughout) automatically passes the horizontal line test, connecting this section directly to Week 6 §6.6: monotonic functions are exactly the functions guaranteed to have an inverse.
11.5 The Derivative of an Inverse Function
If f is differentiable and one-to-one with f′(f−1(a)) ≠ 0, then f−1 is differentiable at a, and
The reflection across y=x from §11.4 explains why: reflecting a tangent line across y=x swaps its rise and run, which inverts its slope — the slope of f−1 at a point is the reciprocal of the slope of f at the corresponding reflected point.
Example 11.3 — Derivative of an inverse function
Let f(x) = x3+2x−1. Given that f(1)=2, find (f−1)′(2) without solving for f−1 explicitly.
Step 1 — Identify a=f(1)=2, so f⁻¹(2)=1
By Eq. 11.3, since f(1)=2, it must be that f−1(2)=1.
Step 2 — Find f'(x) and evaluate at x=1
f′(x) = 3x2+2, so f′(1) = 5.
Step 3 — Apply Eq. 11.4
(f−1)′(2) = 1/f′(f−1(2)) = 1/f′(1) = 1/5
11.6 The Natural Logarithm, Defined as an Integral
Week 9 §9.5 left a gap: ∫ x−1 dx has no power-rule antiderivative. This course fills the gap by defining a new function as that integral:
By FTC Part 1 (Week 9 §9.1), differentiating this definition immediately gives
— which finally completes Eq. 9.5a's antiderivative table:
11.7 Properties of the Natural Logarithm
Because d/dx[ln x] = 1/x > 0 for x > 0, ln x is increasing on its whole domain (Week 6 §6.6), hence one-to-one, hence invertible (§11.4) — its inverse is defined in §11.8. The algebraic properties familiar from precalculus all follow from the integral definition:
| Property | Statement |
|---|---|
| Product | ln(xy) = ln x + ln y |
| Quotient | ln(x/y) = ln x − ln y |
| Power | ln(xr) = r ln x |
| At x=1 | ln 1 = 0 (the integral from 1 to 1 is zero, §9.7) |
11.8 The Exponential Function
Since ln is increasing and one-to-one, it has an inverse, called the natural exponential function exp(x), and the number e is defined as exp(1), i.e. the unique number with ln e=1. Writing exp(x) as ex, Eq. 11.3 gives
Applying the inverse-function derivative rule (Eq. 11.4) with f(x)=ln x, f′(x)=1/x, at a = ex (so f−1(a) = x):
ex is the unique function (up to a constant multiple) that is its own derivative — and consequently its own antiderivative:
For any base a > 0, ax is defined as ex ln a, and logax = ln x/ln a. Every exponential and logarithmic function, in any base, is therefore built from ex and ln x alone; the Chain Rule (Week 4 §4.10) then differentiates any of them.
Example 11.4 — Differentiating with the Chain Rule
Differentiate y = e3x2 and y = ln(5x+1).
Step 1 — Chain Rule on e raised to an inner function
d/dx[e3x2] = e3x2·6x
Step 2 — Chain Rule on ln of an inner function
d/dx[ln(5x+1)] = 15x+1·5 = 5/(5x+1)
11.9 Worked Capstone Example
Example 11.5 — Integrating with the logarithm and exponential
Evaluate ∫ xex2 dx and ∫ tan x dx.
Step 1 — Substitute for the exponential integral (Week 9, §9.6)
u=x2, du=2x dx: ∫ xex2 dx = (1/2)∫ eu du = (1/2)eu+C
= (1/2)ex2 + C
Step 2 — Rewrite tan x and recognise a log form
tan x = sin x/cos x. Let u=cos x, du=−sin x dx:
∫ sin xcos x dx = −∫ 1u du = −ln|u|+C
∫ tan x dx = −ln|cos x| + C — a new standard integral, made possible only once Eq. 11.7 existed.
11.10 Nomenclature
| Symbol | Meaning |
|---|---|
| f−1 | Inverse function of f |
| one-to-one | No two distinct inputs give the same output |
| ln x | Natural logarithm, defined as ∫_1^x (1/t)dt |
| e | The unique number with ln e = 1; base of the natural exponential |
| ex | Natural exponential function, the inverse of ln x |
11.11 Problems
- Work, variable force. A force of F(x)=10/x2 N acts on an object from x=1 to x=5 m. Find the work done.
- Work, spring. A spring has natural length 1 m and requires 50 N to stretch it to 1.2 m. Find the work to stretch it from 1 m to 1.5 m.
- Inverse existence. Explain why f(x)=x2 has no inverse on (−∞,∞), but does have one on [0,∞).
- Derivative of an inverse. Let f(x)=2x+sin x. Given f(0)=0, find (f−1)′(0).
- Logarithm derivative. Differentiate y = ln(x2+1).
- Logarithm properties. Simplify ln(x3√y) − ln z as a single logarithm's expansion, then reverse the process.
- Exponential derivative. Differentiate y = e−2x sin x using the Product Rule.
- Exponential integral. Evaluate ∫ x2ex3 dx.
- Logarithmic integral. Evaluate ∫1e (1/x) dx, and interpret the answer using Eq. 11.5.
- Combined. Evaluate ∫ cot x dx, following the method of Example 11.5.
11.12 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §6.5 (work and fluid forces), §7.1 (inverse functions and their derivatives), §7.2 (natural logarithms), §7.3 (exponential functions).
- Pearson MyLab Math — the Week 11 homework set is assigned directly from the sections above; see the Calendar for the due date.