Calculus

Calculus is the mathematical study of continuous change. It is a core prerequisite for actuarial mathematics, underpinning Probability Theory, financial mathematics, and life contingencies through differentiation, integration, and series analysis.

Limits and Continuity

A limit describes the value a function approaches as its input approaches a point.

$\lim_{x \to a} f(x) = L$

A function $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$ .

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ , where $L$ and $M$ are real numbers:

Rule	Formula
Sum	$\lim_{x \to a}[f(x) + g(x)] = L + M$
Product	$\lim_{x \to a}[f(x) \cdot g(x)] = L \cdot M$

L'Hôpital's Rule — if $f$ and $g$ are differentiable on an open interval containing $a$ (except possibly at $a$ ) and $\lim_{x \to a} \frac{f(x)}{g(x)}$ has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ , provided the limit on the right exists or is $\pm\infty$ .

Differentiation

The derivative of $f$ at $x$ measures the instantaneous rate of change:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Rule	Formula
Power	$(x^n)' = nx^{n-1}$
Product	$(fg)' = f'g + fg'$
Quotient	$\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$
Chain	$[f(g(x))]' = f'(g(x))\cdot g'(x)$
Exponential	$(e^x)' = e^x, \quad (a^x)' = a^x \ln a$
Logarithm	$(\ln x)' = \dfrac{1}{x}$

Partial derivatives treat all variables except one as constants. They appear when working with joint distributions in Probability Theory:

$\frac{\partial f}{\partial x}(x,y) \quad \text{holds } y \text{ fixed}$

Finding the Force of Mortality The survival function is $S(t) = e^{-\lambda t}$ . Find the force of mortality $\mu(t) = -\dfrac{S'(t)}{S(t)}$ .

Answer $S'(t) = -\lambda e^{-\lambda t}$ $\mu(t) = -\frac{-\lambda e^{-\lambda t}}{e^{-\lambda t}} = \lambda$

Integration

The definite integral gives the net signed area under $f$ from $a$ to $b$ . If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ :

$\int_a^b f(x)\,dx = F(b) - F(a)$

$\text{where } F'(x) = f(x)$

This result is the Fundamental Theorem of Calculus.

Technique	When to Use	Key Formula
Power Rule	Polynomial terms	$\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ , $n \neq -1$ (for $n = -1$ : $\ln\lvert x \rvert + C$ )
Substitution	Composite functions	Let $u = g(x)$ , then $du = g'(x)\,dx$
Integration by Parts	Product of functions	$\int u\,dv = uv - \int v\,du$
Partial Fractions	Rational functions	Decompose denominator into linear and irreducible quadratic factors

Improper integrals over $[0,\infty)$ are essential for continuous distributions on an unbounded support:

$\int_0^\infty f(x)\,dx = \lim_{b\to\infty}\int_0^b f(x)\,dx$

Double integrals appear in joint distribution calculations:

$P(X \leq a,\, Y \leq b) = \int_{-\infty}^a \int_{-\infty}^b f(x,y)\,dy\,dx$

Expected Value of the Exponential Distribution Let $X \sim \text{Exp}(\lambda)$ with $f(x) = \lambda e^{-\lambda x}$ , $x > 0$ . Find $E[X]$ .

Answer Using integration by parts with $u = x$ and $dv = \lambda e^{-\lambda x}\,dx$ : $E[X] = \int_0^\infty x\lambda e^{-\lambda x}\,dx = \left[-xe^{-\lambda x}\right]_0^\infty + \int_0^\infty e^{-\lambda x}\,dx = 0 + \frac{1}{\lambda} = \frac{1}{\lambda}$

Series

A series is the sum of the terms of a sequence. Two families appear constantly in actuarial work.

Geometric Series — forms the basis of present-value annuity formulas:

$\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}, \quad |r| < 1$

Taylor / Maclaurin Series — approximates a function as an infinite polynomial near $a = 0$ :

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\,x^n$

Function	Maclaurin Series
$e^x$	$\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$
$\ln(1+x)$	$\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}, \quad -1 < x \leq 1$
$(1+x)^k$	$\displaystyle\sum_{n=0}^{\infty} \binom{k}{n} x^n, \quad \lvert x \rvert < 1$

Geometric Series: Present Value of a Perpetuity An annuity pays $ $1$ at the end of each year forever at effective annual rate $i$ . Find the present value.

Answer Let $v = \dfrac{1}{1+i}$ . The present value is: $PV = v + v^2 + v^3 + \cdots = \sum_{k=1}^{\infty} v^k = \frac{v}{1-v} = \frac{1}{i}$