TITLE: Other forms of the Lagrange theorem?
QUESTION [0 upvotes]: I found a form of the lagrange theorem that I don't know,
I didn't find something similar on the internet.

Suppose that $f:[a,b]\times[c,d] \subset \mathbb{R}^2 \longrightarrow \mathbb{R}$ is a continuous function. Consider the function
\begin{align}
I:[c,d] &\longrightarrow \, \mathbb{R}\\
y\,\,\, &\longmapsto\, I(y) = \int_{a}^{b} f(x,y)\, dx\\
\end{align}

Then they say:

Take a y $\in $ [c,d]. From the Lagrange theorem it follows that $\exists \,\theta \in (0,1) $ so that: 
  \begin{align}
\frac{I(y+k)-I(y)}{k} &= \int_a^b \frac{f(x,y+k)-f(x,y)}{k} \, dx \\
&= \int_a^b \frac{\partial f}{\partial y}(x,y+\theta k) \, dx
\end{align}

The only form of the lagrange theorem that I know is this one:
$$ f'(c) = \frac{f(b)-f(a)}{b-a}. $$
Does someone know how they get an angle theta?

REPLY [2 votes]: It's actually the same thing in different notation: if we write $f(x,y) = F_x(y)$, then the quotient inside the integral is
$$ \frac{F_x(y+k)-F_x(y)}{k}. $$
By Lagrange (AKA the mean value theorem), there is a $c$ with $y<c<y+k$ so
$$ \frac{F_x(y+k)-F_x(y)}{k} = F_x'(c). $$
Any such $c$ can be written as $(1-\theta)y+\theta (y+k) = y + \theta k$ for some $0<\theta<1$, and this is what $\theta $ is: a real number between $0$ and $1$, which depends on $y,k$ and $F_x$.
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