Prove that the antiderivative of an integrable function is both bounded and integrable

Let $f: [a,b] \to \mathbb{R}$ be a bounded function which is also integrable. Define $F: [a,b] \to \mathbb{R}$ by

$$F(x)=\int_{a}^xf(t)\ dt$$

To prove that $F(x)$ is also bounded and integrable

I was able to show that $F(x)$ $\Leftarrow$ the integral of a constant $M$ which is the bound of $f(x)$, but I'm not sure if that's the right way to go about this problem and if it is, I'm not sure where to go from there. Hints and help would be appreciated!

Hint: let $M$ be the bound of $|f|$. You get $$\left|\int_a^b F(x) dx\right|= \left|\int_a^b \int_a^x f(y)dy dx \right|\leq \int_a^b\int_a^x |f(y)|dy\leq \int_a^b M(b-a) dx=M(b-a)^2$$
• Double integral must appear in some way, since you have it once in the definition of $F$, then for the integrability of $F$... – Milly Nov 24 '14 at 20:53