## Integration of partial derivatives

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

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

 Quote by 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.
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.

Recognitions:
Gold Member
Staff Emeritus

## Integration of partial derivatives

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.

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

 Then does this hold? I = F(x,v) - F(x,u) ∂I/∂x = f(x,v) - f(x,u)
 Recognitions: Gold Member Science Advisor Staff Emeritus 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 $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, $$\frac{\partial}{\partial v}F(x, v)= f(x,v)$$ We can write [itex]F(x,u)= \int_u^a f(x,t)dt= -\int_a^u f(x,t)dt$ 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.

 Tags integral derivatives