Let $f:[0,4] \in R$ be a continuous function such that $|f(x)| \leq 2$ for all $x \in[0,4]$ and $\int\limits_0^4 f(t) d t=2$. Then, for all $x \in[0,4]$ the value of $\int\limits_0^x f(t) d t$ lies in the interval

$[-6+2 x, 10-2 x]$

Let $\phi(x)=\int\limits_0^x f(t) d t$. Then, $\phi(x)$ is continuous on $[0,4]$ and is differentiable on $(0,4)$ such that $\phi'(x)=f(x)$. Applying Lagrange's mean value Theorem to $\phi(x)$ on $[x, 4]$, we get

$\phi'(c) =\frac{\phi(4)-\phi(x)}{4-x}$ for some $c \in(x, 4)$

$\Rightarrow f(c) =\frac{2-\phi(x)}{4-x}$

$\Rightarrow |f(c)|=\left|\frac{2-\phi(x)}{4-x}\right| $

$\Rightarrow |f(c)|=\left|\frac{\phi(x)-2}{4-x}\right|$

$\Rightarrow \left|\frac{\phi(x)-2}{4-x}\right| \leq 2$

$\Rightarrow |\phi(x)-2| \leq 8-2 x$

$\Rightarrow -8+2 x \leq \phi(x)-2 \leq 8-2 x$

$\Rightarrow 2 x-6 \leq \phi(x) \leq 10-2 x$

$\Rightarrow \phi(x) \in[2 x-6,10-2 x]$