Integration of partial derivatives

[unscientific]

Homework Statement

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

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)?

Homework Statement

 

[HallsofIvy]

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

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

 

[unscientific]

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

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

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. $\int_u^v f(x,t)dt$ is a function of u, v, and x, NOT t.

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

 

[unscientific]

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. $\int_u^v f(x,t)dt$ is a function of u, v, and x, NOT t.

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

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 $I= \int_{u(x)}^{v(x)}f(x,t)dt$ and then $I= F(x,v)- F(x,u)$ 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.<br /> <br /> By the fundamental theorem of Calculus, <br /> $$\frac{\partial}{\partial v}F(x, v)= f(x,v)$$<br /> <br /> We can write $F(x,u)= \int_u^a f(x,t)dt= -\int_a^u f(x,t)dt$ so that <br /> $$\frac{\partial}{\partial u}F(x,u)= -f(x,u)$$[/itex]$$<br /> Now, to find the derivative <b>with respect to x</b> use the chain rule.$$[/itex]