Prove that if the integral of a function from the same point to two different points is equal, then the function has a root. If there exist $c,d\in[a,b]$ such that $c\neq d$ and $\int\limits_a^c f(x)dx =\int\limits_a^d f(x)dx$ then prove that there exists $t\in(a,b)$ such that $f(t)=0$ (there are further assumptions - see below)
In the first part of the problem, we assume that $f$ is continuous in [a,b]. I solved the problem by using the mean value theorem for integrals and showing that $\int\limits_c^d f(x)dx=0$ (assuming $c < d$).
Now we assume that $f$ is bounded in $[a,b]$ and continuous in $(a,b)$. I intuitively understand that the problem is essentially the same as in the first part, as the $t$ I found in the first part is neither $a$ nor $b$, and the basic shape of the function (and thus the value of the integral) does not change as the result of the changed assumptions - I am having trouble formalizing the proof, however. 
I'd appreciate any help, and I'd especially appreciate general advice on how to translate intuitive understanding to a formal mathematical proof.
Edit: In case it wasn't clear, the assumptions have changed and we now assume that $f$ is bounded in $[a,b]$ and continuous in $(a,b)$. My proof for the first part is longer valid because that MVT only applies when $f$ is continuous in $[a,b]$.
 A: Without loss of generality $d>c:$
$$\int_a^c f(x)\;dx=\int_a^d f(x)\;dx$$
$$\int_a^c f(x)\;dx=\int_a^c f(x)\;dx+\int_c^d f(x)\;dx$$
$$\int_c^d f(x)\;dx=0$$
Since $f$ is bounded, for $x\in (c,d):$
$$\inf_{x\in(c,d)} f\leq f(x)\leq \sup_{x\in(c,d)} f$$
$$(d-c)\inf_{x\in(c,d)} f\leq \int_c^d f(x)dx\leq (d-c)\sup_{x\in(c,d)} f$$
$$\inf_{x\in(c,d)} f\leq 0\leq \sup_{x\in(c,d)} f$$
Since $f$ is continuous there must exist $\lambda\in (c,d)$ such that $f(\lambda)=0$
A: First, I'll remark that the Mean Value Theorem for integrals indeed holds when $f$ is bounded and assumed to be continuous only on $(a,b)$:
Claim:  Let $f$ be bounded on $[a,b]$ and continuous on $(a,b)$.  Then there is a number $c\in(a,b)$ such that $f(c)={1\over b-a}\int_a^b f(x)\,dx$.

Proof: Define the function $F(x)=\int_a^x f(t)\,dt$. Since $f$ is integrable over $[a,b]$, $F$ is defined on $[a,b]$. It follows from the Fundamental Theorem of Calculus that $F$ is continuous on $[a,b]$. Since $f$ is continuous at any $c\in(a,b)$, it follows, again from the Fundamental Theorem, that $F'(c)$ exists for $c\in(a,b)$ and that for such points  $F'(c)=f(c)$.
So $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$.
We may thus apply the Mean Value Theorem for derivatives to the function $F$ over the interval $[a,b]$. Doing so provides a $c\in(a,b)$ such that
$$
{F(b)-F(a)\over b-a}=F'(c).
$$ 
But, as $c\in(a,b)$, $F'(c)=f(c)$. Moreover $F(a)=0$, and $F(b)=\int_a^b f(x)\,dx$. 
The result follows.



Secondly, I'll remark that you don't need this theorem to prove your result: 
Assume $f$ is bounded, continuous on $(a,b)$ and that $\int_c^d f(x)\,dx=0$ with $d>c$.  
Assume, in the hope of obtaining a contradiction, that $f$ does not assume the value $0$ in $(a,b)$. Then since $f$ is continuous on $(a,b)$, it must either be positive over $(a,b)$ or negative over $(a,b)$.  Without loss of generality, assume $f(x)>0$ for all $x\in(a,b)$.  
Choose $\alpha\in(c,d)$.
Since $f$ is continuous at $\alpha$, there is a nondegenerate  interval $I\subset [c,d]$ with $\alpha\in I$ and $f(x)>f(\alpha)/2>0$ for all $x\in I$.  Now define the function $g:[c,d]\rightarrow \Bbb R$ via $g(x)=f(\alpha)/2$ for $x\in I$ and $g(x)=0$ for $x\in[c,d]\setminus I$.
We have $g(x)< f(x)$ for all $x\in[c,d]$.  Moreover, $g$ is integrable over $[c,d]$ and
$$
0< {f(\alpha)\over 2}\cdot|I|=\int_c^d g(x)\,dx\le\int_c^df(x)\,dx; 
$$
a contradiction.
A: Why use the MVT for integrals in the first part? 
Why not just the observation that $\int_{a}^{c}f(x) dx + \int_{c}^{d}f(x) dx = \int_{a}^{d}f(x)dx ?$
