1. False.    [0,1], f(x):=x, g(x):=x²

^{2. True.  If  f is cont. on [a,b], then}  

^{is an antiderivative, F'(z)=f(z)}

Perhaps you are interested in

residue(f, x=a)

see help page.

Hope it helps.

                            Sandor

 

 

Hi Robert,

Could you give a link and/or a  subject?

Reading your answer I absolutely don't  know which one maybe the previous one.

Thanks,     Sandor

Thanks.

The solution of Mariner in this special case is obviously the best.

For other not-builtin types John's solution will  be useful, so I  memorize  both of them.

Sandor

Thanks for your answers.

I always learn a lot of useful things.

My problem was to accept symbolic constant in a procedure so I use the solution of Robert;

smth:=proc(T)
local ispositive;
ispositive:=rcurry(is,positive);
if type(T,satisfies(ispositive))
   then 1
   else 0
end if;
end proc;

the suggestion of Acer is very exciting, but I have to learn more about new type.

Sandor

 

 

Thanks for all. I wanted to solve this problem in a similar way as Robert. (I'm thinking of a modification that works well. If I find a solution I will post it here.) At this moment jakubi's solution is the simplest (that works also if integral is not separated). Anyway, I learned some useful methods from you. Thanks again. Sandor

(Very sorry, something is wrong. I use a lot of script, it can be the reason.) I want to (triple)integrate f(x,y,z):=7z on the cylinder

x^2+y^2 less than R, 0 less than z less than m.
<\pre>

There are a lot of predefined region, but I haven't found Cylinder.
In fact, my question is not "how to solve this exercise", but
"does exist in Maple somewhere in a package where I can define the region Cylindrical which I could use in integration.
In VectorCalculus there is SetCoordinates('cylindrical'[r, theta,z]); but I don't know
it can be applied and how, or not.

Thanks, Sandor

I want to (triple)integrate f(x,y,z):=7z on the cylinder x^2+y^2<=R, 0<=z<=m. There are a lot of predefined region, but I haven't found Cylinder. Could you help me? Thanks, Sandor

Hello, A := array(1..10); Digits:=4: for i from 1 to 10 etc.

I used pre and /pre, but it was unable to show. Sorry.

value(J);
a less than -1, undefined,

otherwise 

                     /                     /            (1/2)  \\
                     |                     |    / 2    \       ||
                     |                     |    |a  - 1|       ||
                     |                   ln|a + |------|      a||
                     |                     |    |   2  |       ||
                     |                     \    \  a   /       /|
                     |                   -----------------------|
                     |                                 (1/2)    |
                     |                         / 2    \         |
                     |                         |a  - 1|         |
                     |                       a |------|         |
                     |                         |   2  |         |
                     \                         \  a   /         /

Why a less than -1 undefined?

Probably I caused a bit confusion. Sorry. Thanks for your valuable answers. Acer, is right. I meant expression (which defines a math function). Joe, the indeterminate name is not known, it is the choice of a user. Here I consider only one-variable (math) functions (=expressions). Later, maybe, I want to consider vector or multi-variable (math) functions (=expressions). And I want to allow the constant math functions, also, so if Pi gets returned, then it is good. I will investigate your suggestions. Acer, "It's not known whether the procedure will do evaluations of the expression at many points" . This is the situation. Knowing these, what is your advice?