An integration inequality 
$f$ is differentiable on [-1,1], $M=\sup|f'|$. There is $a \in (0,1)$ such that $\int_{-a}^a f(x)dx=0$. Prove that
  $$
\left|\int^1_{-1}f(x)dx\right| \le M(1-a^2)
$$

I don't know how to use $\int^a_{-a} f(x) dx = 0$.

$$
\left| \int_{-1}^1 f(x) \mathrm{d}x \right|=
\left|\int_{-1}^{-a} f(x) \mathrm{d}x  + \int_{a}^1 f(x) \mathrm{d}x\right| \\
\le \int_{-1}^{-a} |f(x)| \, \mathrm{d}x + \int_a^1 |f(x)| \, \mathrm{d}x.
$$
As mentioned in the comments, $\int_{-a}^a f(x)\mathrm{d}x=0$ implies that $f(x)=0$ for some $x\in(-a,a)$, since if $f(x)$ was either strictly positive or strictly negative, then its integral over $(−a,a)$ could not be $0$. Let $x_0\in(-a,a)$ be the value of $x$ where $f(x_0)=0$. Then, as $\left\lvert f'(x) \right\rvert \le M$, we have that for every $x\in[-1,1]$,
$$|f(x)| =|f(x)-f(x_0)| \le M|x-x_0|.$$
Therefore
$$\int_{-1}^{-a} |f(x)| \, \mathrm{d}x\le M\int_{-1}^{-a} (x_0-x) \, \mathrm{d}x=M\int_a^1(x_0+x)\mathrm{d}x,$$
$$\int_{a}^{1} |f(x)| \, \mathrm{d}x\le M\int_{a}^{1} (x-x_0) \, \mathrm{d}x,$$
which gives us the bound
$$M\int_a^1(x_0+x)\mathrm{d}x+M\int_{a}^{1} (x-x_0) \, \mathrm{d}x=M\int_a^1 2x\mathrm{d}x=M(1-a^2).$$
