# Definite integral with variable limits of a multivariable function.

1. Mar 17, 2013

### crocomut

I have the following integral:

$\int_0^{f(x,y)}{f' \sin(y-f')df'}$

Now suppose that f(x,y) = x*y, my question is how do I write the integral in terms of x and y only? Can I do something like this?

Since $df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$ we can obtain:

$\int_0^{x y}{x' y' \sin(y'-x' y') (\frac{\partial f'}{\partial x'}dx'+\frac{\partial f'}{\partial y'}dy'})$

2. Mar 18, 2013

### HallsofIvy

First, you should be careful to use different letters for variables in the integrand and in the limits of integration. So write the integral as
$$\int_0^{f(x,y)} f'(u,v) sin(v- f'(u,v)) d(f'(uv))$$

Now, what do you mean by f' where f is a function of two variables? The derivative with respect to some parameter, t, so that $df= (\partial f)(\partial x)(dx/dt)+ (\partial f)(\partial y)(dy/dt)$? (NOT df'- that would involve two derivatives).

First, of course, if f= xy, then f' is NOT x'y'. By the product rule, f'= xy'+ x'y. And "y" in the sine would not automatically become y'. The integral would be
$$\int_0^{x(t)y(t)}(x(s)y'(s)+ x'(s)y(s))sin(y(s)+ x'(s)y(s)+ x(s)y'(s))(f_x(x(s),y(s)) x'(s)+ f_y(x(s),y(s))y'(s)) ds$$

3. Mar 18, 2013

### crocomut

Thanks for your reply, I apologize for the confusion but I don't think I was clear, the apostrophe was meant to indicate different letters for variables, not a derivative. With that in mind, I will rephrase my question and actually simplify the integrand since that is not what is important.

Suppose we have $$\int^{f(x,y)}_0{g(u,v)dg}$$. Given $$f(x,y) = xy$$ and $$g(u,v) = uv$$, now since $$dg=\frac{\partial g}{\partial u}du+\frac{\partial g}{\partial v}dv = vdu + udv$$, is the integral then equal to:

$$\int^{f(x,y)}_0{g(u,v)dg} = \int^{xy}_0{u v^2 du} + \int^{xy}_0{u^2 v dv}$$

Last edited: Mar 18, 2013
4. Mar 18, 2013

### HallsofIvy

What you have written makes no sense. You can certainly have $$\int g dg$$ but "g" now is a dummy variable- you cannot then assert that g= uv.

What is true is that $$\int_0^{xy} g dg= \left[\frac{1}{2}g^2\right]_0^{xy}= \frac{1}{2}x^2y^2$$.

5. Mar 18, 2013

### crocomut

Excellent. I knew it didn't make sense and just needed confirmation, I got the integral from an obscure paper on a computational code and that confirms what I suspected - the integral is not written properly. Thanks.

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