Integration, Area & Triple Integrals — Maths for Chemistry

Integration is differentiation run backwards

It measures accumulated change — and the area under a curve

Where differentiation finds a rate, integration finds a total. Geometrically the definite integral is the area under the curve between two limits.

\int_{a}^{b} f (x) d x = Δ x \to 0 lim i \sum f (x_{i}) Δ x = F (b) - F (a)

The Fundamental Theorem ties it to the antiderivative $F$ : integrate the power $x^{n}$ and you raise the power and divide, $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ — the exact reverse of the power rule.

The standard integrals to keep at hand

Power (n ≠ −1)

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C

Reciprocal → log

\int \frac{1}{x} d x = ln ∣ x ∣ + C

Exponential

\int e^{k x} d x = \frac{1}{k} e^{k x} + C

Sine & cosine

\int cos x d x = sin x + C

What chemists integrate

Integrate a rate law and you get the integrated rate law (the straight-line plots). Integrate p over V and you get work. Integrate

∣ ψ ∣^{2}

and you get probability — the condition that normalises a wavefunction. Try the Riemann slicer to build the area from scratch, then work the sample problems in each section.

Common pitfalls

A definite integral subtracts both limits. $\int_{a}^{b} f d x = F (b) - F (a)$ — evaluating only at the top limit is the most frequent error.
The reciprocal breaks the power rule. $\int x^{- 1} d x = ln ∣ x ∣$ , not $x^{0} /0$ . This single exception is where every logarithm in thermodynamics comes from.
Indefinite integrals need $+ C$ . In chemistry that constant is fixed by an initial condition (e.g. $[A] = [A]_{0}$ at $t = 0$ ) — omit it and the integrated rate law loses its intercept.
Change the limits when you substitute. A $u$ -substitution in a definite integral must convert the $x$ -limits into $u$ -limits (or back-substitute before evaluating).

Riemann slicer

Build the area under a curve from rectangles, then watch it converge to the exact definite integral as the slices get thinner.

Add rectangles and watch the estimate close in on the exact area

\int_{0}^{2} x^{2} d x

number of rectangles, n6

Riemann sum2.6481

exact integral2.6667

error0.0185

Each rectangle is height × width = $f (x_{i}) Δ x$ . Summing them estimates the area; the limit as $Δ x \to 0$ is the definite integral.

Worked example 1A definite integral by the power rule

Evaluate $\int_{0}^{3} x^{2} d x$ exactly.

Worked example 2Area under a rate is an amount

Early in a synthesis the rate of formation of NH₃ is roughly constant at $3.0 \times 1 0^{- 3} mol L^{- 1} s^{- 1}$ for the first 20 s. Estimate the concentration of NH₃ formed.

p–V work lab

The work of an isothermal expansion is the area under the p–V curve — and integrating 1/V is exactly where the logarithm comes from.

The shaded area under p–V is the work of an isothermal expansion

W = \int_{V_{1}}^{V_{2}} \frac{n RT}{V} d V = n RT ln \frac{V _{2}}{V _{1}}

final volume V₂ (V₁ = 1 L)4.0 L

expansion ratio V₂/V₁4.00

work done by gas3435 J

The area (an integral) and the formula nRT ln(V₂/V₁) give the same number — that's the integral being evaluated. Integrating 1/V is what produces the logarithm.

Worked example 3Work of a reversible isothermal expansion

2.00 mol of an ideal gas expands isothermally and reversibly at 300 K from 5.0 L to 15.0 L. Calculate the work done by the gas. ( $R = 8.314 J K^{- 1} mo l^{- 1}$ )

Worked example 4Where the formula comes from

Derive $w = n RT ln (V_{2} / V_{1})$ by integrating the ideal-gas pressure $p = n RT / V$ over the volume change.

Triple integrals & normalization

Stretch the definite integral into 3D: integrating |ψ|² over all space is the condition that normalises a wavefunction.

A triple integral over all space normalises the hydrogen 1s orbital

∭ ∣ ψ ∣^{2} r^{2} sin θ d r d θ d ϕ = \int_{0}^{\infty} P (r) 4 π r^{2} ∣ ψ ∣^{2} d r = 1

integrate out to radius R2.50 a₀

electron density within R87.5%

most-probable radius1.00 a₀

whole-space integral1 (normalised)

The triple integral runs over r, θ and φ. The two angular integrals contribute a constant 4π, collapsing it to a single radial integral of P(r) = 4πr²|ψ|². The shaded area is the probability of finding the electron inside R; push R → ∞ and it reaches exactly 1. The peak sits at the Bohr radius a₀.

Worked example 5Normalising the particle in a box

The ground state of a particle in a 1-D box is $ψ (x) = N sin \frac{π x}{L}$ for $0 \leq x \leq L$ . Find the normalisation constant $N$ .

Worked examples

Integration as the engine behind the integrated rate laws and total yields.

Worked example 6Deriving the first-order integrated rate law

Starting from $\frac{d [ A ]}{d t} = - k [A]$ , integrate to obtain the integrated first-order rate law.

Worked example 7Total product from an integrated rate

A reaction releases gas at a rate $r (t) = 0.50 e^{- 0.10 t} mol s^{- 1}$ . How much gas is produced in total (as $t \to \infty$ )?

ChallengeEntropy change on heating with a temperature-dependent heat capacity

The molar heat capacity of a solid is $C_{p} = a + b T$ with $a = 22.6 J K^{- 1} mo l^{- 1}$ and $b = 0.0100 J K^{- 2} mo l^{- 1}$ . Find $Δ S$ when one mole is heated reversibly from 300 K to 400 K.

Check yourself

Six questions linking the area, the antiderivative and the chemistry.

Question 1 of 6 · Score 0

Integration is the reverse process of:

Choose an answer.