Prove 2-variables integral is continuous 
Let $f(t)$ be a differentiable function on $[a,b]$. Let ${ F(x,y) = \int_{x}^{y}{f(t)\,dt} }$. Show that $F(x,y$) is continuous in the rectangle $[a,b] \times [a,b]$.

I think that it should be proved by continuity definition, but I'm not sure how to approach it... Any help? 
 A: Hint: Consider first a special case. Try to prove using $\varepsilon$ and $\delta$ that the function $G:[c,d] \to \Bbb{R}$ defined by
\begin{equation}
G(\xi) = \int_a^{\xi} f(t) \, dt
\end{equation}
is continuous. Then, define the projection maps $\pi_1: (x,y) \mapsto x$ and $\pi_2 (x,y) \mapsto y$. With this, the map $F$ you are interested in can be written as a difference of composite maps:
\begin{align}
F(x,y) &= \int_x^y f(t) \, dt \\
&= \int_a^y f(t) \, dt - \int_a^x f(t) \, dt \\
&= (G \circ \pi_2)(x,y) - (G \circ \pi_1)(x,y)
\end{align}
Using continuity of $\pi_1,\pi_2$, can you see how to conclude that $F$ is continuous?

Side remark: If you know that $f$ is differentiable, you can conclude a lot more about $F$. By the fundamental theorem of calculus, $G$ is twice differentiable; hence $F$ will be as well.
To prove that $F$ is continuous, the only assumption you need is that $f$ is Riemann integrable on $[a,b]$, because using the fact $f$ is bounded, you can prove $G$ is Lipschitz continuous on $[a,b]$, which implies uniform continuity and hence continuity of $G$. So, my point is even if you assume the bare minimum about $f$, once you integrate it, you get a much nicer function.
A: The assumption is too strong. it is enough if $f$ is continuous or integrable. The function $g(x)=\int _a^{x}f(t)\, dt$ is a continuous and $F(x,y)=g(y)-g(x)$. This is a difference of two continuous functions on $[a,b] \times [a,b]$ and hence it is continuous. 
For an $\epsilon -\delta$ proof note that $|g(x)-g(x')| <\epsilon$ if $|x-x'| <\frac {\epsilon} M$ where $M=\sup \{|f(x)|:a\leq x \leq b\}$.  From this it should be easy to complete the proof. 
A: I think that you can use the theorem of total differential:
By fundamental theorem of calculus for every $x$ fixed $\partial _yF(x,y)=f(y)$ is continuos with respect $y$ while for every $y$ fixed $\partial_xF(x,y)=-f(x)$ that is continuos then $F$ is a differentiable function and so it must be continuos. 
