Theory

Integral calculus and economic applications

How to find the antiderivatives of a function with basic integrals, by substitution, by parts and with partial fractions, what the definite integral and the integral function are, what the mean value theorem and the fundamental theorem of calculus say, and how integrals are used to compute areas, total costs, surpluses and present values.

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Antiderivatives and the indefinite integral
Basic integrals

Antiderivatives and the indefinite integral

1. Antiderivative

Differentiation assigns to a function its slope; here we look for the inverse operation, that is, a function whose derivative is known. Let $f$ be defined on an interval $I$ . A function $F$ differentiable on $I$ is an antiderivative of $f$ on $I$ if

F'(x) = f(x) \qquad \text{for every } x \in I .

For example $F(x) = x^3$ is an antiderivative of $f(x) = 3x^2$ on $\mathbb{R}$ , because $\left( x^3 \right)' = 3x^2$ . The functions $x^3 + 5$ and $x^3 - 2$ are also antiderivatives of $3x^2$ , because the derivative of a constant is zero; a function that has an antiderivative therefore has infinitely many.

2. Antiderivatives of the same function

Let $F$ be an antiderivative of $f$ on the interval $I$ . Then the antiderivatives of $f$ on $I$ are all and only the functions

F(x) + c, \qquad c \in \mathbb{R} .

Every $F + c$ is an antiderivative, because $(F + c)' = F' = f$ . Conversely, if $G$ is another antiderivative, the difference $G - F$ has derivative $f - f = 0$ on the whole interval, and by the theorem on the sign of the derivative and monotonicity it is constant. The graphs of the antiderivatives are vertical translations of one another, and the constant is fixed by assigning the value at one point. The antiderivative of $2x$ that equals $3$ at $x = 1$ is $x^2 + c$ with $1 + c = 3$ , that is, $x^2 + 2$ .

y = x^2

y = x^2 + 2

Two antiderivatives of

2 x

; the graphs of

x^{2}

and of

x^{2} + 2

are

2

apart vertically for every

x

, and at every abscissa they have the same slope; the point

(1, 3)

singles out

x^{2} + 2

3. Indefinite integral

The indefinite integral of $f$ is the set of all its antiderivatives, and it is written

\int f(x)\,dx = F(x) + c ,

where $F$ is any antiderivative and $c$ is an arbitrary constant. The function $f$ is called the integrand, and $dx$ indicates the variable with respect to which we integrate. From the differentiation rules for the sum and for the product by a constant follows linearity: for $h, k \in \mathbb{R}$

\int \big( h\,f(x) + k\,g(x) \big)\,dx = h \int f(x)\,dx + k \int g(x)\,dx .

For example $\displaystyle\int \left( 3x^2 - 4x + 5 \right) dx = x^3 - 2x^2 + 5x + c$ , as one checks by differentiating the result. The check by differentiation works for every indefinite integral.

Basic integrals

Reading the table of derivatives of the elementary functions backward gives, on the intervals where the functions are defined,

$f(x)$	$\displaystyle\int f(x)\,dx$
$x^{a}$ , with $a \neq -1$	$\dfrac{x^{a + 1}}{a + 1} + c$
$\dfrac{1}{x}$	$\ln \lvert x \rvert + c$
$e^{x}$	$e^{x} + c$
$a^{x}$ , with $a > 0$ and $a \neq 1$	$\dfrac{a^{x}}{\ln a} + c$
$\sin x$	$-\cos x + c$
$\cos x$	$\sin x + c$

The absolute value in the antiderivative of $\dfrac{1}{x}$ is needed for $x < 0$ ; indeed for $x < 0$ we have $\big( \ln(-x) \big)' = \dfrac{-1}{-x} = \dfrac{1}{x}$ . The power formula includes roots and reciprocals, which are first written as powers:

\int \left( \sqrt{x} + \frac{1}{x^2} \right) dx = \int \left( x^{\frac{1}{2}} + x^{-2} \right) dx = \frac{2}{3}\,x^{\frac{3}{2}} - \frac{1}{x} + c .

Basic integrals in composite form

Let $g$ be differentiable on an interval. From the chain rule follow

\int g(x)^{a}\,g'(x)\,dx = \frac{g(x)^{a + 1}}{a + 1} + c \quad (a \neq -1), \qquad \int \frac{g'(x)}{g(x)}\,dx = \ln \lvert g(x) \rvert + c, \qquad \int e^{g(x)}\,g'(x)\,dx = e^{g(x)} + c .

They are recognized when the integrand contains a function $g$ and, as a factor, its derivative $g'$ , possibly up to a multiplicative constant, which is adjusted. For example

\int \frac{2x}{x^2 + 3}\,dx = \ln\left( x^2 + 3 \right) + c, \qquad \int x \left( x^2 + 1 \right)^3 dx = \frac{1}{2} \int 2x \left( x^2 + 1 \right)^3 dx = \frac{\left( x^2 + 1 \right)^4}{8} + c ,

and the absolute value is omitted in the first case because $x^2 + 3 > 0$ . In the second integral the factor $2$ of the derivative of $x^2 + 1$ is missing, so we multiply and divide by $2$ . With $g(x) = 3x$ we also obtain $\displaystyle\int e^{3x}\,dx = \dfrac{e^{3x}}{3} + c$ .

The theory continues with 7 more sections (integration by substitution and by parts, integrals of rational functions, the definite integral…), plus all worked examples, the solutions and the PDF.

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