Fundamental Theorem of Calculus for definite integrals… assume continuity?

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 '14 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 '14 at 5:36
Here you can find a doc explaining why it isn't possible in general blog.wolfram.com/2008/01/19/… – Dr. belisarius Mar 20 '14 at 3:30
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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]
*)

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FToC? what is that? – Hector Mar 20 '14 at 3:55
@Hector Fundamental Theorem of Calculus ...sorry lazy writer – Dr. belisarius Mar 20 '14 at 3:59

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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It'd be nice to know why those are irreducible, or what Mathematica needs to know about $u(x,y)$ to get the desired result. – jvriesem Aug 3 '15 at 22:37