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So here's the problem:

I can evaluate the indefinite integral:

Integrate[D[u[x], x], x]
u[x]

However, I'd like to evaluate:

Integrate[D[u[x],x], {x, x0, x1}]

and get

u[x1] - u[x0]

Or especially, evaluate

Integrate[D[u[x, y], x], {x, x0, x1}]

and get

u[x1, y] - u[x0, y]

Is there a way that I can assume that D[u[x], x] is continuous in the range x0 to x1? Is there a some assumption that can be met in order for me to evaluate the fundamental theorem of calculus?

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Sorry, Integral=Integrate. I've edited that. That may work, but specifically, I'm trying to evaluate: Integrate[ D[u[x, y], x] + D[u[x, y], y], {x,x0,x1}] Or even: Integrate[ D[u[x, y], x,x] + D[u[x, y], y,y], {x,x0,x1}, {x,x0,x1}] –  Jon Feb 17 at 5:28
    
That may work, but specifically, I'm trying to evaluate: Integrate[ D[u[x, y], x] + D[u[x, y], y], {x,x0,x1},{y,y0,y1}] Or even: Integrate[ D[u[x, y], x,x] + D[u[x, y], y,y], {x,x0,x1}, {x,x0,x1},{y,y0,y1},{y,y0,y1}] However, I think your approach may work. –  Jon Feb 17 at 5:36
1  
Here you can find a doc explaining why it isn't possible in general blog.wolfram.com/2008/01/19/… –  belisarius Mar 20 at 3:30
    
Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! –  belisarius Mar 20 at 17:36

2 Answers 2

The problem is that Mathematica can't guess that your functions are analytical and so complying with the hypotheses of the FToC.

An equivalent situation is this:

Limit[f[x], x -> x0]
(*
Limit[f[x], x -> x0]
*)

But:

Limit[f[x], x -> x0, Analytic -> True]
(*
f[x0]
*)

So you could use the ability of the Limit[] function to understand when a function is analytic to use the FToC as follows (sorry, it's trivial anyway):

k[x_] := Integrate[D[u[x], x], x]
Limit[k[x], x -> x1, Analytic -> True] -  Limit[k[x], x -> x0, Analytic -> True]
(*
-u[x0] + u[x1]
*)
share|improve this answer
    
FToC? what is that? –  Hector Mar 20 at 3:55
    
@Hector Fundamental Theorem of Calculus ...sorry lazy writer –  belisarius Mar 20 at 3:59

From your comment

I'm trying to evaluate: Integrate[ D[u[x, y], x] + D[u[x, y], y], {x,x0,x1},{y,y0,y1}]

and if I understand correctly, this is kind of a negative answer.

If you plug this into Mathematica (v 9.0.1)

Integrate[D[u[x, y], x] + D[u[x, y], y], x, y]

you get this

$$\int u(x,y) \, dx+\int u(x,y) \, dy$$

which shows you that there are irreducible integrals there.

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