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Probably my question is related to this question, but this question does not provide the answer to my question.

Let $f:X\times Y \rightarrow \mathbb{R}$. Suppose I am interested in integration of $f(x,y)$ over $x\in X$. But problem is that I know $x$ and $y$ are related through variable $z$. For example, $x=g(z)$ and $y=h(z)$. In this case, usually I need to take the relationship between $x$ and $y$ in consideration when I integrate. But is there any way to integrate $f(x,y)$ over $x$ ignoring relationship between $x$ and $y$ just like partial derivative?

user1292919
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  • To the other linked question:when you have 2-variable functions, you can do double integrals. 3-variable functions you and do triple integrals. single integrals of multivariable functions are precisely "partial integration". If you understand partial differentiation (single derivatives) you understand single integrals of multivariable functions. For instance f(x,y) means that for each $y$, you have a single variable function. Therefore$\partial f/\partial x$gives you the derivative graph of all thoses single variable functions.Thats a lot of derivatives all packed into$\partial f/\partial x$ – DWade64 Oct 15 '18 at 00:25
  • Likewise $\int f(x,y)dx$ gives you the antiderivative of all those single variable functions (1 single variable function for each $y$). That's a lot of integrals of single variable functions all packed into $\int f(x,y)dx$ – DWade64 Oct 15 '18 at 00:27

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Integrating the function $f$ over one of it's inputs is just as valid as differentiating with respect to one of it's inputs. However $f$ and the function $f(g(z), h(z))$ are very different functions. In fact, they are 2 completely different functions. $f$ is the way it's defined. $f(g(z), h(z))$ is a composition of functions and therefore it's a totally different function from $f$. It's a single variable function of $z$ (over a totally different domain than $f$). You can either

$$ \int f(x,y) dx$$

Or equivalently integrate over the parameterized path $x = g(z)$

$$\int f(g(z), y) g'(z) dz $$

Or integrate the completely different composition function $J(z) := f(g(z), h(z))$

$$\int f(g(z), h(z))dz = \int J(z) dz $$

DWade64
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  • The second integration $\int f(g(z),y)g'(z)dz= \int f(g(z),h(z))g'(z)dz$ right? or can I ignore $y$? – user1292919 Oct 15 '18 at 03:19
  • @user1292919 The 2 integrals that you wrote down are not the same. The first varies $z$, which will ends up only varying $x$. $y$ is held constant in this integral. The second varies $z$, but doesn't make a lot of sense.The $x$ and $y$ components of $f$ will vary in this second integral, but you have a weird $g'(z)$ which, now, no longer has an interpretation – DWade64 Oct 15 '18 at 10:07
  • If you want to integrate the composite function (a totally different function than $f$), then do $\int f(g(z), h(z)) dz$. If you want to integrate with $y$ held fixed, then do $\int f(g(z),y) g'(z) dz$ – DWade64 Oct 15 '18 at 10:08