# Definite integral versus indefinite integral evaluation

Why evaluating $$\iint x\, \mathrm dx\, \mathrm dx$$ in $[0;2]$ is different from calculating $$\int^2_0 \int^2_0 x\, \mathrm dx\, \mathrm dx$$ ?

What is the conceptual difference between the two?

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The answers themselves are different: $$\iint xdxdx=\dfrac{x^3}{6}+C$$ $$\int^2_0\int^2_0 xdxdx=16$$Other than that, your question is unclear. – Sanath K. Devalapurkar May 20 '14 at 15:56
Did you mean $[0,2]^2$ for the first integral? – 7raiden7 May 20 '14 at 16:02
@SanathDevalapurkar The solution to the definite integral is 4. – Peter Woolfitt May 20 '14 at 16:02
@PeterWoolfitt Yes, it is. My mistake! – Sanath K. Devalapurkar May 20 '14 at 16:03
Edited, it is more clear? – mattecapu May 20 '14 at 16:05

Let's simplify a bit the two expression:

1. $$\int_0^2\int x\, \mathrm dx\, \mathrm dx=\int_0^2 \frac{x^2}{2} +C \, \mathrm dx$$
2. $$\int^2_0 \int^2_0 x\, \mathrm dx\, \mathrm dx = \int_0^2\frac{2^2}{2}\, \mathrm dx=\int_0^22\, \mathrm dx$$

So simplifying both of them, you'll notice that you're evaluating in the same interval $[0,2]$, two different functions!

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Now I understand. Specifically, when integrating acceleration to get space, we find first the $s(t)$ function by doing $\iint a(t)\,dt\,dt$ or we do the double definite integral? And what is the meaning of the second integral I wrote? Does it make sense to integrate over a number (not a constant function)? – mattecapu May 20 '14 at 18:48
1) I suggest you visit this page:hyperphysics.phy-astr.gsu.edu/hbase/acons.html 2) Of course! It makes sense! You can look at it as the means to calculate the area of a rectangle of height $2$ and base $2$. Generally speaking $\int_0^xk dx$ can be seen as the value of the area of a rectangle whose dimensions are $x$ and $k$. – sirfoga May 20 '14 at 19:02
That confirm my thought. Thank you. – mattecapu May 20 '14 at 19:05
You're welcome! ;) – sirfoga May 20 '14 at 19:05

The problem lies in your notation. You shouldn't use a variable both inside and outside the integral. Physicists do usually write stuff like $\int_0^t f(t)dt$ but mathematically you should write $\int_0^t f(t')dt'$. The integration variable exists only inside the integral sign.

For indefinite integral, it's sort of confusing to write multiple integrals... neither notation seems right. In that case one should at least write definite integrals with only one boundary (absend lower boundary accounts for the indeterminate integration constant), like $$\int^x \int^{x'}x''\,dx''\,dx'$$ however that's not a common notation. I guess $\iint x\,dx\,dx$ is sort of acceptable. Still, my expression above tells you where the difference lies. The limits of your two integrals are fundamentally different. The indefinite integral "telescopes" in a sense that the upper limit of the inner integral is used as a variable for the outer integral. Your particular choice of definite integral has a fixed interval in both directions and actually separates into a product of two unrelated integrals.

For the second integral, you should definitely write

$$\int_0^2\int_0^2 x\,dx\,dy$$

I clears things up and shows you that the outer integral really doesn't have anything to do with the $x$ in the inner integral.

Teachers in high school really should stress very strongly that the naming of the integration variable is totally arbitrary. When instead of $\int_0^1 x\,dx$ you write $\int_0^1 Д\,dД$ people do start paying attention more. Especially if you use symbols that kids in your country are ignorant about. They don't even have to be from a real language. Just make up some squigglies. Let them know that it's just a dummy variable that has no meaning outside the integral.

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I don't think you got my point, my problem isn't variabile notation/scope. – mattecapu May 20 '14 at 18:46
@mattecapu Well rewriting the indefinite integral with the upper limits solved your question and you can't do that without cleaning up the notation. As you see your definite integral is a number - zero free parameters. And the indefinite integral leaves the boundary variable, so you acutally get a function of x. It's a completely different kind of object. The first one has a square domain. The second is triangular (draw the limits). – orion May 20 '14 at 20:29
I agree, but notation is just a side aspect of the problem to me – mattecapu May 20 '14 at 20:36
Fine, ignore that part, I got carried away. – orion May 20 '14 at 20:37