Integral Calculus

Calculation Rules for undefined Integrals

a) Additivity A sum under the integral is integrated by the summands are integrated separately and then summed.

$\int (f (x) + g (x)) dx = \int f (x) dx + \int g (x) dx$

b) Factor Rule A constant factor a can be pulled out of the integral.

$\int a f (x) dx = a \int f (x) dx$

c) If F(u) is the antiderivative of f(u), then the following relationship holds for arbitrary constants a and b with a not equal to 0.

$\int f (a x + b) dx = \frac{1}{a} F (a x + b) + C$

d) If f(x) is differentiable and f(x) not equal to 0, then the following relationship applies.

$\int \frac{f^{'} (x)}{f (x)} dx = \ln | f (x) | + C$

e) Integration by parts Are the functions u(x) and v(x) are differentiable, then the following relationship holds.

$\int (u (x) v^{'} (x)) dx = u (x) v (x) - \int u^{'} (x) v (x) dx$

f) Substitution If the function f(z) is continuous and z = g(x) is differentiable, then the following relationship holds.

$\int (f (g (x)) g^{'} (x)) dx = \int f (z) dz$

Basic Integrals

General

$\begin{array}{l} \int f' (x) dx & = & f (x) \\ \int dx & = & x \end{array}$

Power

$\begin{array}{l} \int x^{n} dx & = & \frac{x^{n+1}}{n+1} \\ \int {(a x + b)}^{n} dx & = & \frac{{(a x + b)}^{n+1}}{a (n + 1)} \\ \int \frac{dx}{x} & = & \ln | x | \end{array}$

Exponential Functions

$\begin{array}{l} \int e^{x} dx & = & e^{x} \\ \int a^{x} dx & = & \frac{a^{x}}{\ln a} \end{array}$

Trigonometric Functions

$\begin{array}{l} \int \sin x dx & = & - \cos x \\ \int \cos x dx & = & \sin x \\ \int \tan x dx & = & - \ln | \cos x | \\ \int \frac{dx}{\cos^{2} x} & = & \tan x \\ \int \frac{dx}{\sin^{2} x} & = & - \cot x \end{array}$

Logarithmic Functions

$\begin{array}{l} \int \ln x dx & = & x \ln x - x \\ \int {(\ln x)}^{2} dx & = & x {(\ln x)}^{2} - 2 x \ln x + 2 x \\ \int \frac{dx}{x \ln x} & = & \ln (\ln x) \end{array}$

Irrational Functions

$\begin{array}{l} \int \sqrt{a x + b} dx & = & \frac{2}{3 a} \sqrt{{(a x + b)}^{3}} \\ \int \frac{dx}{\sqrt{a x + b}} & = & \frac{2 \sqrt{a x + b}}{a} \end{array}$

Examples

Example: Linearity of the integral (Rule a and b)

$\int (5 x + 2 \sin (x)) dx$

$= \int 5 x dx + \int 2 \sin (x) dx$

Due to the additivity of the summands can be integrated individually.

$= 5 \int x dx + 2 \int \sin (x) dx$

The constant factors are taken outside the integral.

$= \frac{5}{2} x^{2} - 2 \cos (x) + C$

With the basic integrals, the solution follows.

Example: A function of a linear function (Rule c)

$\int \frac{1}{2 x + 3} dx$

$f (a x + b) = \frac{1}{2 x + 3}$

The function under the integral is a function of a linear function.

$F (u) = \int \frac{1}{u} du = \ln | u | + C$

Formation of the primitive of f with the substitution u = ax + b.

$= \frac{1}{2} \ln | 2 x + 3 | + C$

Install in accordance with rule c yields the solution of the integral.

Example: f(x) in the denominator and the derivative of f(x) in the nominator (Rule d)

$\int \frac{\cos (x)}{\sin (x)} dx$

$\frac{d}{dx} \sin (x) = \cos (x)$

That means in the numerator is the derivative of the denominator.

$= \ln | \sin (x) | + C$

Install in accordance with rule d yields the solution of the integral.

Example: Integration by parts (Rule e)

$\int x^{3} \ln (x) dx$

$\int x^{3} dx = \frac{x^{4}}{4}$

Is the first factor v', then the integral of the first factor is v.

$\frac{d}{dx} \ln (x) = \frac{1}{x}$

The derivative of the second factor is u'.

$= \frac{x^{4}}{4} \ln (x) - \frac{1}{4} \int x^{3} dx$

Put in following rule e.

$= \frac{x^{4}}{4} \ln (x) - \frac{1}{4} \frac{x^{4}}{4} + C$

With the integral of I the solution of the integral follows.

$= \frac{x^{4}}{4} (\ln (x) - \frac{1}{4}) + C$

This can be written also shorter by factoring out.

Example: Integration by substitution (Rule f)

$\int \frac{x dx}{\sqrt{a^{2} + x^{2}}}$

$g (x) = \sqrt{a^{2} + x^{2}}$

Substitution of the denominator as a function g(x).

$\frac{dg}{dx} = \frac{x}{\sqrt{a^{2} + x^{2}}} = \frac{x}{g}$

The derivative of g(x) by dx is the relationship according to the differentials.

$x dx = g dg$

Dissolved, the term can be used in the integral.

$= \int \frac{g dg}{g} = \int dg = g + C$

Put in and integrated.

$= g + C = \sqrt{a^{2} + x^{2}} + C$

Repatriation of substitution yields the solution.

Rules of calculation of definite integrals

The existence of the integrals in each case provided. There are a, b, c ∈ ℜ constants.

a) Interchange of the limits of integration.

$\int_{a}^{b} f (x) dx = - \int_{b}^{a} f (x) dx$

a1) Definition

$\int_{a}^{a} f (x) dx = 0$

b) Factor Rule A constant factor a can be pulled outside the integral.

$\int_{a}^{b} a f (x) dx = a \int_{a}^{b} f (x) dx$

c) Sum Rule A sum under the definite integral can be integrated by means of the summands.

$\int_{a}^{b} f (x) + g (x) dx = \int_{a}^{b} f (x) dx + \int_{a}^{b} g (x) dx$

d) Decomposition of the definite integral in part integrals.

$\int_{a}^{b} f (x) dx = \int_{a}^{c} f (x) dx + \int_{c}^{b} f (x) dx$

e) Mean Value Theorem If f is integrable and we have m ≤ f ≤ M, then there exists at least one number μ with m ≤ μ ≤ M and we have the following relationship.

$\int_{a}^{b} f (x) dx = μ (b - a)$

f) General Mean Value Theorem If f and g are integrable and we have m ≤ f ≤ M and g ≥ 0 either always or g ≤ 0, then there exists at least one number μ with m ≤ μ ≤ M and it works as follows.

$\int_{a}^{b} f (x) g (x) dx = μ \int_{a}^{b} g (x) dx$

g) Second Mean Value Theorem If f is monotonic and bounded and g integrable, then there exists at least one number μ for which the following relationship is valid.

$\int_{a}^{b} f (x) g (x) dx = f (a) \int_{a}^{μ} g (x) dx + f (b) \int_{μ}^{b} g (x) dx$

h) If a function f is continuous and differentiable, then applies

$F (x) = \int_{a}^{x} f (t) dt$

with $F' (x) = f (x)$

i) Main Theorem of the Differential and Integral Calculations. If the antiderivative F of f is known, then the definite integral of f is calculated as follows

$\int_{a}^{b} f (x) dx = F (b) - F (a)$

j) Integration by parts Are the functions u (x) and v (x) is differentiable, then the following relationship holds.

$\int_{a}^{b} (u (x) v^{'} (x)) dx = {[u (x) v (x)]}_{a}^{b} - \int_{a}^{b} u^{'} (x) v (x) dx$

k) Substitution for definite integrals.

$\int_{a}^{b} (f (g (x)) g^{'} (x)) dx = \int_{g (a)}^{g (b)} f (z) dz$