Real Analysis: Prove that $|\int_a^b f(x)dx-(b-a)f(a)|<\frac{1}{2}M(b-a)^2$. Suppose $|f'(x)|<M$ for all $x\in\mathbb{R}$. 
Prove that
$|\int_a^b f(x)dx-(b-a)f(a)|<\frac{1}{2}M(b-a)^2$.
What I have so far as thoughts:
I am thinking I will want to use the Mean Value Theorem, but I am lost as to how I need to start? Any hints would be very helpful!

We have 
$$
| \int_a ^b f(x) dx - (b-a)f(a)\,| 
= | \int_a ^b (f(x)-f(a) dx \,|
\leq  \int_a ^b |(f(x)-f(a)\,| dx
$$
By mean value theorem, we have
$$
|(f(x)-f(a)\,| \leq M |x-a| =M (x-a)
$$
Hence 
$$
| \int_a ^b f(x) dx - (b-a)f(a)\,| \leq \int_a ^b M (x-a) dx \leq \frac{1}{2} 
M (b-a)^2
$$
