Integration of partial derivatives

unscientific

1. The problem statement, all variables and given/known data

The problem is attached in the picture. The top part shows what is written in the book, but im not sure how they got to (∂I/∂v)...


3. The attempt at a solution

It's pretty obvious in the final term that the integral is with respect to 't' while the differential is with respect to 'v' . How did they simply convert F(x,v) into f(x,v)?
1. The problem statement, all variables and given/known data

Attached Thumbnails

 

HallsofIvy

You understand that the variable of integration is a "dummy" variable don't you? That the [itex]\int_a^x f(t)dt[/itex] is a function of x, not t.

Here, an example would be
[tex]\int_{t= 3x}^{x^2} t^2- 2x dt= \left[\frac{1}{3}t^3- 2xt\right]_{3x}^{x^2}= \frac{1}{3}x^6- 2x^4- \left(\frac{1}{3}(27x^3)- 6x^2\right)[/tex]
a function of x, not t.

unscientific

Yes, but how can you reverse the integration by ∂/∂v ? Shouldn't it be ∂/∂t instead?

It's like saying F(x,y) = int f(x,y) dy

then

f(x,y) = ∂/∂z F(x,y)


when they are clearly different variables - z and y.

HallsofIvy

If you say "yes" then you are you saying that you understand that this integral is NOT a function of t so it cannot be differentiated with respect to t. Go back and read what I said again. [itex]\int_u^v f(x,t)dt[/itex] is a function of u, v, and x, NOT t.

You are the one who is try to differentiate with an incorrect variable.

unscientific

So we simply look at what's the end-product, F(x,v) instead of the intermediate step?

unscientific

Then does this hold?

I = F(x,v) - F(x,u)

∂I/∂x = f(x,v) - f(x,u)

HallsofIvy

When they say [itex]I= \int_{u(x)}^{v(x)}f(x,t)dt[/itex] and then [itex]I= F(x,v)- F(x,u)[/itex] they are really saying that [itex]F(x,v)= \int_a^{v(x) f(x,t)dt[/tex] and [itex]F(x,u)= \int_u^a f(x,t)dt[/tex] where a is any constant.

By the fundamental theorem of Calculus,
[tex]\frac{\partial}{\partial v}F(x, v)= f(x,v)[/tex]

We can write [itex]F(x,u)= \int_u^a f(x,t)dt= -\int_a^u f(x,t)dt[/itex] so that
[tex]\frac{\partial}{\partial u}F(x,u)= -f(x,u)[/itex]

Now, to find the derivative with respect to x use the chain rule.