Mathematics / Mathematik / Matemática

Mathematics / Mathematik / Matemática

Posts 1-2 of 2

Winfried Weber Group moderator
The company name is only visible to registered members.

Surface integals
Hello,

sorry about my missing knowledge but I would like to calculate the following Integrals:

one-dimensional (is easy):
Integral_I f(x) dx

with I=[0,1] and f(x) = x*x - x + 0,5
(=> [1/3* x^3 - 1/2 * x^2 + 0,5*x ] is the primitive funktion, so i=1/3 is the solution.

two-dimensional:
Integral_I f(x , y) d(x,y)

with I=[0,1] x [0,1]
and f(x, y) = (x*x + y*y +x*y) - (x + y) + 0,5

n-dimensional:

Integral_I f(x1, ... xn) d (x1, ... xn)

with I=[0,1] x [0,1] ... x [0,1] (n-cube) and

f(x1 , ... xn) = Summe_ (i,j ) (xi * xj) - Summe_i (xi) + 0,5

thanks for support

Winfried
- 17 Apr 2008, 8:14 pm
Winfried Weber wrote: > Hello, > > sorry about my missing knowledge but I would like to calculate the > following Integrals: > > one-dimensional (is easy): > Integral_I f(x) dx > > with I=[0,1] and f(x) = x*x - x + 0,5 > (=> [1/3* x^3 - 1/2 * x^2 + 0,5*x ] is the primitive funktion, so > i=1/3 is the solution. > > > two-dimensional: > Integral_I f(x , y) d(x,y) > > with I=[0,1] x [0,1] > and f(x, y) = (x*x + y*y +x*y) - (x + y) + 0,5 > > > n-dimensional: > > Integral_I f(x1, ... xn) d (x1, ... xn) > > with I=[0,1] x [0,1] ... x [0,1] (n-cube) and > > f(x1 , ... xn) = Summe_ (i,j ) (xi * xj) - Summe_i (xi) + 0,5 > > > thanks for support > > Winfried
Post visible to registered members