Ian is correct. BTW,

Operation $f*g$, $(f*g)(x)=\int_{-\infty}^\infty f(x-y)g(y)\,dy$ is called a *convolution* (we implicitly met it and will discuss when study Fourier transform),

$H(x)$ is called Heaviside function and $(H*f)(x)=\int_{-\infty}^x f(y)\,dy$. Therefore the answer is such integral of $Q(x)$ and

\begin{equation*}

(Q*H)(x)=\left\{\begin{aligned}

& 0 \qquad&& x<-1,\\

&x+1 && -1<x<1,\\

&2 && x\ge 1.

\end{aligned}\right.

\end{equation*}