 Integral Calculus

Basic Integrals

General

$\begin{array}{lll}\int f\prime \left(x\right)\mathrm{dx}& =& f\left(x\right)\\ \int \mathrm{dx}& =& x\end{array}$

Power

$\begin{array}{lll}\int {x}^{n}\mathrm{dx}& =& \frac{{x}^{\mathrm{n+1}}}{\mathrm{n+1}}\\ \int {\left(ax+b\right)}^{n}\mathrm{dx}& =& \frac{{\left(ax+b\right)}^{\mathrm{n+1}}}{a\left(n+1\right)}\\ \int \frac{\mathrm{dx}}{x}& =& \mathrm{ln}|x|\end{array}$

Exponential Functions

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

Trigonometric Functions

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

Logarithmic Functions

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

Irrational Functions

$\begin{array}{lll}\int \sqrt{ax+b}\mathrm{dx}& =& \frac{2}{3a}\sqrt{{\left(ax+b\right)}^{3}}\\ \int \frac{\mathrm{dx}}{\sqrt{ax+b}}& =& \frac{2\sqrt{ax+b}}{a}\end{array}$

Calculation Rules for undefined Integrals

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

$\int \left(f\left(x\right)+g\left(x\right)\right)\mathrm{dx}=\int f\left(x\right)\mathrm{dx}+\int g\left(x\right)\mathrm{dx}$

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

$\int af\left(x\right)\mathrm{dx}=a\int f\left(x\right)\mathrm{dx}$

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.

$\int f\left(ax+b\right)\mathrm{dx}=\frac{1}{a}F\left(ax+b\right)+C$

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

$\int \frac{{f}^{\prime }\left(x\right)}{f\left(x\right)}\mathrm{dx}=\mathrm{ln}\left|f\left(x\right)\right|+C$

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

$\int \left(u\left(x\right){v}^{\prime }\left(x\right)\right)\mathrm{dx}=u\left(x\right)v\left(x\right)-\int {u}^{\prime }\left(x\right)v\left(x\right)\mathrm{dx}$

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

$\int \left(f\left(g\left(x\right)\right){g}^{\prime }\left(x\right)\right)\mathrm{dx}=\int f\left(z\right)\mathrm{dz}$

Examples

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

$\int \left(5x+2\mathrm{sin}\left(x\right)\right)\mathrm{dx}$

$=\int 5x\mathrm{dx}+\int 2\mathrm{sin}\left(x\right)\mathrm{dx}$

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

$=5\int x\mathrm{dx}+2\int \mathrm{sin}\left(x\right)\mathrm{dx}$

The constant factors are taken outside the integral.

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

With the basic integrals, the solution follows.

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

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

$fax+b = 12x+3$

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

$Fu=∫1udu=ln|u|+C$

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

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

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)

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

$ddxsinx = cosx$

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

$=\mathrm{ln}|\mathrm{sin}\left(x\right)|+C$

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

Example: Integration by parts (Rule e)

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

$∫x3dx = x44$

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

$ddxlnx = 1x$

The derivative of the second factor is u'.

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

Put in following rule e.

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

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

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

This can be written also shorter by factoring out.

Example: Integration by substitution (Rule f)

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

$gx = a2+x2$

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

$dgdx = xa2+x2 = xg$

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

$xdx = gdg$

Dissolved, the term can be used in the integral.

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

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.

$\underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}=-\underset{b}{\overset{a}{\int }}f\left(x\right)\mathrm{dx}$

a1) Definition

$\underset{a}{\overset{a}{\int }}f\left(x\right)\mathrm{dx}=0$

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

$\underset{a}{\overset{b}{\int }}af\left(x\right)\mathrm{dx}=a\underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}$

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

$\underset{a}{\overset{b}{\int }}f\left(x\right)+g\left(x\right)\mathrm{dx}=\underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}+\underset{a}{\overset{b}{\int }}g\left(x\right)\mathrm{dx}$

d) Decomposition of the definite integral in part integrals.

$\underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}=\underset{a}{\overset{c}{\int }}f\left(x\right)\mathrm{dx}+\underset{c}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}$

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.

$\underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}=\mu \left(b-a\right)$

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.

$\underset{a}{\overset{b}{\int }}f\left(x\right)g\left(x\right)\mathrm{dx}=\mu \underset{a}{\overset{b}{\int }}g\left(x\right)\mathrm{dx}$

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.

$\underset{a}{\overset{b}{\int }}f\left(x\right)g\left(x\right)\mathrm{dx}=f\left(a\right)\underset{a}{\overset{\mu }{\int }}g\left(x\right)\mathrm{dx}+f\left(b\right)\underset{\mu }{\overset{b}{\int }}g\left(x\right)\mathrm{dx}$

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

$F\left(x\right)=\underset{a}{\overset{x}{\int }}f\left(t\right)\mathrm{dt}$

with $F\prime \left(x\right)=f\left(x\right)$

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

$\underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}=F\left(b\right)-F\left(a\right)$

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

$\underset{a}{\overset{b}{\int }}\left(u\left(x\right){v}^{\prime }\left(x\right)\right)\mathrm{dx}={\left[u\left(x\right)v\left(x\right)\right]}_{a}^{b}-\underset{a}{\overset{b}{\int }}{u}^{\prime }\left(x\right)v\left(x\right)\mathrm{dx}$

k) Substitution for definite integrals.

$\underset{a}{\overset{b}{\int }}\left(f\left(g\left(x\right)\right){g}^{\prime }\left(x\right)\right)\mathrm{dx}=\underset{g\left(a\right)}{\overset{g\left(b\right)}{\int }}f\left(z\right)\mathrm{dz}$