The integral — from rates back to amounts
You know the reaction rate at every instant:
$$\text{Rate}(t) = k[A]_0 e^{-kt}$$
Question: How much product formed between t = 0 and t = 10 seconds?
The rate tells you how fast — but you want how much.
You need to add up all the infinitesimal amounts formed at each instant.
That's the integral.
The derivative takes amounts → rates.
The integral takes rates → amounts.
They are inverse operations.
| Derivative | Integral |
|---|---|
| Position → Velocity | Velocity → Position |
| Concentration → Rate | Rate → Concentration change |
| Energy → Force | Force → Work |
| Charge → Current | Current → Charge |
| Situation | Integral gives you |
|---|---|
| "How much total?" | ∫ rate dt |
| "Area under this curve?" | ∫ f(x) dx |
| "Work done by varying force" | ∫ F dx |
| "Total probability" | ∫ P(x) dx |
| "Average value" | (1/(b-a)) ∫ f(x) dx |
How do you find the area under $y = f(x)$ from $x = a$ to $x = b$?
Strategy: Approximate with rectangles, then take the limit as rectangles become infinitely thin.
The Definite Integral: As n → ∞ (rectangles become infinitely thin):
$$\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x = \int_a^b f(x) \, dx$$
If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$, then:
$$\int_a^b f(x) \, dx = F(b) - F(a)$$
Integration and differentiation are inverse operations.
To evaluate a definite integral:
No need for Riemann sums!
$$\frac{d}{dx}\left[\int_a^x f(t) \, dt\right] = f(x)$$
The derivative of the "area so far" function is the original function.
| f(x) | ∫f(x)dx | Notes |
|---|---|---|
| $x^n$ (n ≠ -1) | $\frac{x^{n+1}}{n+1} + C$ | Power rule |
| $x^{-1} = 1/x$ | $\ln|x| + C$ | Special case |
| $e^x$ | $e^x + C$ | |
| $a^x$ | $\frac{a^x}{\ln a} + C$ | |
| $\sin x$ | $-\cos x + C$ | |
| $\cos x$ | $\sin x + C$ |
For reversible expansion of a gas:
$$W = -\int_{V_1}^{V_2} P \, dV$$
$P = nRT/V$ (constant T):
$$W = -\int_{V_1}^{V_2} \frac{nRT}{V} \, dV = -nRT \ln\frac{V_2}{V_1}$$
For a first-order reaction A → B with rate $= k[A]_0 e^{-kt}$:
$$\Delta[B] = \int_0^t k[A]_0 e^{-kt} \, dt = [A]_0(1 - e^{-kt})$$
As t → ∞, all A converts to B: Δ[B] = [A]₀
The most important integral in science:
$$\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$$
$$\int_{-\infty}^{\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}$$
This integral appears everywhere:
u-Substitution: If the integrand has form $f(g(x))g'(x)$, let $u = g(x)$:
$$\int f(g(x)) g'(x) \, dx = \int f(u) \, du$$
Evaluate $\int 2x \cos(x^2) \, dx$
Let $u = x^2$, then $du = 2x\, dx$.
$$\int 2x \cos(x^2) \, dx = \int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C$$
Check: $\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x$ ✓
When limits are infinite:
$$\int_a^{\infty} f(x) \, dx = \lim_{b \to \infty} \int_a^b f(x) \, dx$$
$$\int_1^{\infty} \frac{1}{x^p} \, dx \begin{cases} \text{converges} & \text{if } p > 1 \\ \text{diverges} & \text{if } p \leq 1 \end{cases}$$
| Integral | Result |
|---|---|
| $\int_1^{\infty} \frac{1}{x^2} dx$ | = 1 (converges) |
| $\int_1^{\infty} \frac{1}{x} dx$ | = ∞ (diverges) |
| $\int_0^{\infty} e^{-2x} dx$ | = 1/2 (converges) |
| Concept | Formula |
|---|---|
| FTC | $\int_a^b f(x)dx = F(b) - F(a)$ |
| Power rule | $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ |
| Substitution | $\int f(g(x))g'(x)dx = \int f(u)du$ |
| Average value | $\langle f \rangle = \frac{1}{b-a}\int_a^b f(x)dx$ |
| Gaussian | $\int e^{-ax^2}dx = \sqrt{\pi/a}$ |
| Application | Integral |
|---|---|
| Work (PV) | $W = -\int P\,dV$ |
| Heat | $q = \int C_P \, dT$ |
| Product formed | $\Delta[B] = \int \text{rate}\, dt$ |
| Normalization | $\int P(x)\,dx = 1$ |
| Expectation | $\langle x \rangle = \int x P(x)\,dx$ |