Integration Essentials (for Probability and ML)

The GATE DA syllabus lists only limits, continuity, differentiability, Taylor series and optimisation under calculus. But every continuous probability question — normalising a PDF, computing a CDF, an expectation or a variance — is an integration problem. This page collects exactly what you need.

Antiderivatives and the fundamental theorem

$F$ is an antiderivative of $f$ if $F' = f$. The fundamental theorem of calculus:

1. If $F(x) = \int_a^x f(t)\,dt$ and $f$ is continuous, then $F'(x) = f(x)$. More generally $\frac{d}{dx}\int_{a}^{g(x)}f(t)\,dt = f(g(x))g'(x)$.
2. $\int_a^b f(x)\,dx = F(b) - F(a)$.

Part 1 is the PDF–CDF relationship: $F_X(x) = \int_{-\infty}^x f_X(t)\,dt$ and $F_X' = f_X$.

(all up to a constant $C$)

Techniques

Substitution (reverse chain rule): $\int f(g(x))g'(x)\,dx = \int f(u)\,du$ with $u = g(x)$; change the limits too.

Integration by parts (reverse product rule): $\int u\,dv = uv - \int v\,du$. Choose $u$ to simplify when differentiated (LIATE order: logs, inverse trig, algebraic, trig, exponential).

Symmetry: for odd $f$, $\int_{-a}^a f = 0$; for even $f$, $\int_{-a}^af = 2\int_0^af$. So $E[Z] = \int z\phi(z)\,dz = 0$ for the standard normal without any calculation.

Definite integrals as areas and averages

$\int_a^b f(x)\,dx$ is the signed area under the curve. The average value of $f$ on $[a, b]$ is $\frac{1}{b - a}\int_a^bf$ — an expectation under the uniform distribution.

In probability: $P(a \le X \le b) = \int_a^b f$, $E[g(X)] = \int g(x)f(x)\,dx$, and normalising constants make $\int f = 1$.

Improper integrals

Integrals over infinite ranges or through singularities are limits:

$$\int_a^\infty f(x)\,dx = \lim_{b\to\infty}\int_a^bf(x)\,dx.$$

Comparison test: if $0 \le f \le g$ and $\int g$ converges, so does $\int f$. Heavy-tailed distributions are exactly those where moment integrals like $\int x^2f(x)\,dx$ diverge (the Cauchy distribution has no mean).

Two integrals worth memorising

Gamma function: $\Gamma(\alpha) = \int_0^\infty x^{\alpha - 1}e^{-x}\,dx$, with $\Gamma(n) = (n - 1)!$, $\Gamma(\alpha + 1) = \alpha\Gamma(\alpha)$ and $\Gamma(\tfrac12) = \sqrt\pi$. Consequently

$$\int_0^\infty x^ne^{-\lambda x}\,dx = \frac{n!}{\lambda^{n+1}},$$

which gives all moments of the exponential distribution: $E[X^n] = n!/\lambda^n$.

Gaussian integral: $\int_{-\infty}^\infty e^{-x^2}dx = \sqrt\pi$, hence $\int_{-\infty}^{\infty}e^{-(x - \mu)^2/(2\sigma^2)}dx = \sigma\sqrt{2\pi}$ — the normal density's normalising constant.

Double integrals (for joint densities)

For a joint density over a region, integrate one variable at a time (Fubini), with limits read off the region:

$$P(X < Y) = \int_0^1\int_x^1 f(x, y)\,dy\,dx \quad\text{for } (X, Y)\text{ on the unit square}.$$

Numerical integration

Monte Carlo integration estimates $\int g(x)f(x)\,dx = E[g(X)]$ by averaging $g$ over samples from $f$ — the basis of sampling-based inference.

Quick revision