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If we define $g:\mathbb R \rightarrow \mathbb R$ by $g(x)=2x$, then we can interpret e.g. $\int ^5_2g(x)dx$ as an area.

Now suppose we define $f:\mathbb R \rightarrow \mathbb R^2$ by $f(x)=(2x,3x^2)$. Is there any similar geometric interpretation of $\int ^5_2f(x)dx=(\int ^5_2 2x dx,\int ^5_2 3x^2 dx)$ as a volume/area?

If there's any such geometric interpretation, then what would the volume/area be in the figure below (drawn in CalcPlot3D)?

enter image description here

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    Uhm.. How do you calculate $\int_{2}^{5}(2x,3x^{2}){\rm d}x$? What's the definition? –  Dec 15 '21 at 06:26
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    There is no one fixed interpretation. The interpretation depends on the situation. For example, if I think of $f$ as describing the velocity vector of a particle as it undergoes motion in the plane, then $\int_2^5f$ can be interpreted as the change in position of the particle. See Visualization of double integrals for what I mean by "interpretation depends on the situation". Sure, in that link, I mainly address the interpretation of different domains, while you're asking about the target spaces, but the main message stays the same. – peek-a-boo Dec 15 '21 at 06:29
  • or of course, the most straight forward interpretation is to think of the $(x,y,z)$ coordinate system, and think of the function $f$ as giving two functions $f_1,f_2:\Bbb{R}\to\Bbb{R}$, and think of $f_1$ as "living on the $y$-axis" and $f_2$ as "living on the $z$-axis". Then, $\int_2^5f(x),dx$ can be thought of as the vector consisting of the appropriate areas bounded by the graphs of $f_1$ and $f_2$. This interpretation may sound obvious, but that just goes to show that there is no one fixed interpretation of integrals. – peek-a-boo Dec 15 '21 at 06:33
  • the vector consisting of the appropriate areas --- I'm not sure what you mean by this? I thought a vector is a 1D object (like a line), while an area is a 2D object. If so, then how can a vector consist of areas? –  Dec 15 '21 at 06:37
  • I think you're misinterpreting my sentence. All I meant is that a vector in $\Bbb{R}^2$ means an ordered tuple of numbers. So, $\int_2^5f= \left(\int_2^5f_1, \int_2^5f_2\right)$ is a tuple of numbers. The first entry of the tuple is the number $\int_2^5f_1$, which one can think of as the signed area bounded by the graph of $f_1$ (which you draw on the $x$-$y$ plane) and the $x$-axis from $x=2$ to $x=5$. The second entry of the tuple $\int_2^5f_2$, which again can be thought of as the area bounded by the graph of $f_2$ in such and such region. Hence, we have a vector whose components are areas. – peek-a-boo Dec 15 '21 at 06:42

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