Proof of expression with integrals I have had trouble proving the following expression. Do you have any hints to help me?
Let $f:[a,b]$ be an integrable function for which $$\int_a^bf(x)dx=6$$ Prove that there exist $t_1,t_2\in(a,b)$ such that $$\int_{t_1}^{t_2}f(x)dx=2$$ 
 A: Since $f$ is integrable, the function $F(x)=\int_0^xf(x)\,dx$ is continuous. From this,  the linearity of the integral, and the fact that $F(b)=6$, it follows that there is a $t_1$ with $b>t_1>a$ such that $\int_{t_1}^b f(x)\,dx>2$.
Now let $G(x)=\int_{t_1}^x f(x)\,dx$. $G$ is continuous. Also, we have $G(t_1)=0$ and $G(b)>2$. 
Now, you can wrap things up using the Intermediate Value Theorem.
A: These two cases give a contradiction:


*

*$\forall t_1,t_2\in(a,b)$:


$$F(t_1,t_2)=\int_{t_1}^{t_2}f(x)dx<2$$
so by continuity we get
$$6=\lim_{(t_1,t_2)\to(a,b)}F(t_1,t_2)\le2$$


*

*$\forall t_1,t_2\in(a,b)$:


$$F(t_1,t_2)=\int_{t_1}^{t_2}f(x)dx>2$$
so we get
$$0=\lim_{(t_1,t_2)\to(a,a)}F(t_1,t_2)\ge2$$
so there's  $t'_1,t''_1,t'_2,t''_2$ such that $F(t'_1,t'_2)<2$ and $F(t''_1,t''_2)>2$.
Now we have
$$(a,b)\times (a,b)\ne \{(x,y)\mid F(x,y)<2\}\cup\{(x,y)\mid F(x,y)>2\}$$
using the fact that $(a,b)\times (a,b)$ is a connect set and that the two given set in the union are not the empty set then there's $(t_1,t_2)$ such that $F(t_1,t_2)=2$.
