Assume $f$ is Riemann integrable on $[a,b],$ and that $F'=f$ on $[a,b].$ Using Fundamental theorem of integral calculus, show that $\int_a^b f(x)dx=(b-a)f\{a+\theta(b-a)\}$ holds for some $\theta\in(0,1)$.
I am using the "Fundamental theorem of integral calculus" as
Let $f: [a,b] \to \mathbb{R}$ be a Riemann integrable function. If $F: [a,b] \to \mathbb{R}$ is an antiderivative of $f$, then $$\int_a^b \! f(x) \, \mathrm{d}x = F(b)-F(a).$$
How to use this theorem to show the above?
On
It sounds as though you're trying to prove this for Riemann integrable $f$. It's false in that context: Define $f:[0,2]\to\Bbb R$ by $$f(t)=\begin{cases}1&(0\le t<1),\\-1&(1\le t\le 2).\end{cases}$$Then $\int_0^2 f(t)=0$ although there is no $t$ with $f(t)=0$.
There must be another hypothesis in the problem that you haven't told us about...
On
Consider the function $F$ as you mentioned. What you want to show translates by the Fundamental theorem to:
$F(b)-F(a)=(b-a) \cdot f\big(a+ \theta(b-a) \big)$
Which again rewrites as:
$\dfrac{F(b)-F(a)}{b-a}=f\big(a+ \theta(b-a) \big)$
If you know that $f$ is continuous, then $F$ is differentiable on $(a,b)$ and it's derivative is $f$ . By Lagrange's theorem there exists $c\in (a,b)$ such that:
$\dfrac{F(b)-F(a)}{b-a}= F'(c)$
But $c$ can be written as $a+\theta(b-a)$, for some $\theta \in (0,1)$.
I assume the problem was supposed to be this: Assume $f$ is Riemann integrable on $[a,b],$ and that $F'=f$ on $[a,b].$ Then the conclusion follows as stated.
Proof: Let $I = \int_a^b f.$ Define
$$G(x) = F(x)-\frac{I}{b-a}x,\,\,\,x\in [a,b].$$
Note that $G'(x) = f(x) - \dfrac{I}{b-a}$ on $[a,b].$ Also note $G(b) - G(a) = 0,$ here using the version of the FTC you stated. Thus by the mean value theorem, $G'(c) = 0$ for some $c\in (a,b).$ This implies
$$0=G'(c) = f(c)-\frac{I}{b-a} \, \implies\, I = (b-a)f(c).$$
Since $c,$ like any point in $(a,b),$ can be writen as $a+\theta(b-a)$ for some $\theta \in (0,1),$ we're done.