Theory

Integral calculus

Antiderivatives and integration techniques, the Riemann definite integral with the fundamental theorem of calculus, and the computation of areas.

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Antiderivatives and indefinite integral
The table of elementary integrals

Antiderivatives and indefinite integral

1. Antiderivative and indefinite integral

Let $f : I \to R$ with $I$ an interval. A function $F : I \to R$ is an antiderivative of $f$ on $I$ if $F$ is differentiable and $F^{'} (x) = f (x)$ for every $x \in I$ . The indefinite integral of $f$ is the set of all antiderivatives of $f$ and is denoted by $\int f (x) d x$ . For example $\frac{x ^{3}}{3}$ is an antiderivative of $x^{2}$ , because its derivative is $x^{2}$ . Finding antiderivatives is the inverse operation of differentiation; together with the fundamental formula of calculus it reduces the computation of definite integrals to finding an antiderivative.

2. Two antiderivatives differ by a constant

If $F$ and $G$ are antiderivatives of $f$ on the interval $I$ , there exists $c \in R$ such that $G (x) = F (x) + c$ for every $x \in I$ . Therefore, once an antiderivative $F$ is found, the indefinite integral is described by the expression

\int f (x) d x = F (x) + c, c \in R .

y = \dfrac{x^{3}}{3}

y = \dfrac{x^{3}}{3} + 1

y = \dfrac{x^{3}}{3} - 1

Three antiderivatives of

x^{2}

\frac{x ^{3}}{3}

in color,

\frac{x ^{3}}{3} + 1

dashed,

\frac{x ^{3}}{3} - 1

in red; each graph is a vertical translation of another

Proof

Let $H = G - F$ . The function $H$ is differentiable on $I$ and $H^{'} (x) = G^{'} (x) - F^{'} (x) = f (x) - f (x) = 0$ for every $x \in I$ . Let $x_{1} < x_{2}$ be two points of $I$ . Since $I$ is an interval, $[x_{1}, x_{2}] \subseteq I$ , therefore $H$ is continuous on $[x_{1}, x_{2}]$ and differentiable on $(x_{1}, x_{2})$ and Lagrange’s mean value theorem applies: there exists $ξ \in (x_{1}, x_{2})$ such that

H (x_{2}) - H (x_{1}) = H^{'} (ξ) (x_{2} - x_{1}) = 0.

Hence $H$ takes the same value at every pair of points of $I$ , that is, it is constant: $H (x) = c$ for every $x \in I$ , from which $G (x) = F (x) + c$ .

Remark

The hypothesis that the domain is an interval cannot be removed. The function $\frac{1}{x}$ has domain $(- \infty, 0) \cup (0, + \infty)$ and its antiderivatives are $ln ∣ x ∣ + c_{1}$ for $x > 0$ and $ln ∣ x ∣ + c_{2}$ for $x < 0$ , with two independent constants. The expression $\int \frac{1}{x} d x = ln ∣ x ∣ + c$ implies that one works on one of the two intervals.

y = \ln \left(\left|x\right|\right)

An antiderivative of

\frac{1}{x}

(in color): it equals

ln ∣ x ∣ + 1

for

x > 0

and

ln ∣ x ∣ - 1

for

x < 0

; the dashed one is

ln ∣ x ∣

The table of elementary integrals

1. Elementary integrals

From the table of derivatives, read backward:

\int x^{α} d x = \frac{x ^{α + 1}}{α + 1} + c (α \neq = - 1), \int \frac{1}{x} d x = ln ∣ x ∣ + c, \int e^{x} d x = e^{x} + c, \int a^{x} d x = \frac{a ^{x}}{ln a} + c (a > 0, a \neq = 1), \int sin x d x = - cos x + c, \int cos x d x = sin x + c, \int \frac{1}{cos ^{2} x} d x = tan x + c, \int \frac{1}{1 + x ^{2}} d x = arctan x + c, \int \frac{1}{1 - x ^{2}} d x = arcsin x + c .

Every formula is verified by differentiating the right-hand side.

2. Elementary integrals in composite form

If a function $f (x)$ appears in place of $x$ and the factor $f^{'} (x)$ appears in the integrand, the chain rule, read backward, gives

\int f^{'} (x) [f (x)]^{α} d x = \frac{[ f ( x ) ] ^{α + 1}}{α + 1} + c (α \neq = - 1), \int \frac{f ^{'} ( x )}{f ( x )} d x = ln ∣ f (x) ∣ + c, \int f^{'} (x) e^{f (x)} d x = e^{f (x)} + c, \int f^{'} (x) cos f (x) d x = sin f (x) + c, \int \frac{f ^{'} ( x )}{1 + [ f ( x ) ] ^{2}} d x = arctan f (x) + c .

These integrals are called almost immediate: one recognizes in the integrand the derivative $f^{'} (x)$ , possibly up to a multiplicative constant.

Example

\int tan x d x = - \int \frac{- sin x}{cos x} d x = - ln ∣ cos x ∣ + c, \int cos x sin^{2} x d x = \frac{sin ^{3} x}{3} + c

In the first integral the numerator $- sin x$ is the derivative of the denominator $cos x$ , therefore the formula for $\frac{f ^{'}}{f}$ applies. In the second the factor $cos x$ is the derivative of $sin x$ , therefore the formula for $f^{'} f^{α}$ with $α = 2$ applies.

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

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