Computing Discrete Convolution in terms of unit step function Consider :
$$
x[n]:=\begin{cases}
1&amp;\text{if $3\leq n\leq 8$}\\
0&amp;\text{if otherwise}
\end{cases}
\quad\text{and}\quad
h[n]:=\begin{cases}
1&amp;\text{if $4\leq n \leq 15$}\\
0&amp;\text{if otherwise}
\end{cases}
$$
I wish to compute the convolution $\displaystyle y[n]=(x*h)[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]$. Because I dislike the method of graphing, I wanted to take another approach using unit step function :
$$
x[n]=u[n-3]-u[n-8]
\quad
\text{and}
\quad
h[n]=u[n-4]-u[n-15]
$$
By using distributivity property of the discrete-time convolution, we get :
\begin{align*}
(x*h)[n]&amp;=(u[n-3]-u[n-8])*(u[n-4]-u[n-15])\\ \\
&amp;=\underbrace{u[n-3]*u[n-4]}_{(y_{1}[n])} -\underbrace{u[n-3]*u[n-15]}_{(y_{2}[n])}-\underbrace{u[n-8]*u[n-4]}_{(y_{3}[n])}+\underbrace{u[n-8]*u[n-15]}_{(y_{4}[n])}
\end{align*}
Therefore we have four convolutions for which we can see that :
$y_{1}[n]$ = $\begin{cases}\displaystyle\sum_{3}^{n-4}1&amp;\text{if $7\leq n \leq 11$}\\0&amp;\text{if otherwise}\end{cases}=\begin{cases}n-6&amp;\text{if $7\leq n \leq 11$}\\0&amp;\text{if otherwise}\end{cases}$
$y_{2}[n] = \begin{cases}\displaystyle\sum_{3}^{n-15}1&amp;\text{if $18\leq n \leq 33$}\\0&amp;\text{if otherwise}\end{cases}=\begin{cases}n-17&amp;\text{if $18\leq n \leq 33$}\\0&amp;\text{if otherwise}\end{cases}$
$y_{3}[n] =\begin{cases}\displaystyle\sum_{8}^{n-4}1&amp;\text{if $12\leq n \leq 16$}\\0&amp;\text{if otherwise}\end{cases}=\begin{cases}n-11&amp;\text{if $12\leq n \leq 16$}\\0&amp;\text{if otherwise}\end{cases}$
$y_{4}[n]=\begin{cases}\displaystyle\sum_{8}^{n-15}1&amp;\text{if $23\leq n \leq 38$}\\0&amp;\text{if otherwise}\end{cases}=\begin{cases}n-22&amp;\text{if $23\leq n \leq 38$}\\0&amp;\text{if otherwise}\end{cases}$
I must note that I am afraid I don't know if the upper bound in the sum of $(1)$, $(2)$, $(3)$, and $(4)$ are right or wrong. I think what's left to do is compute :
$$
y[n]=y_{1}[n]-y_{2}[n]-y_{3}[n]+y_{4}[n]
$$
I hope someone can show me how from here we can prove that the answer is :
$$
y[n]=\begin{cases}
n-6&amp;\text{if $7\leq n\leq 11$}\\
6&amp;\text{if $12\leq n\leq 18$}\\
24-n&amp;\text{if $19\leq n \leq 23$}\\
0&amp;\text{if otherwise}
\end{cases}
$$
As a first remark, I would really suggest to also familiarize yourself with the graphical way of convolving two functions or sequences. This will also make your calculations a lot easier because you get some intuition on the nature of the solution.
Second, your definitions of $x[n]$ and $h[n]$ in terms of the step sequence $u[n]$ are not correct. E.g., the given $x[n]$ equals $1$ for $n=8$, but $u[n-3]-u[n-8]$ is zero for $n=8$.
It is formally correct to represent the solution as the sum of four individual convolutions of shifted step sequences, so what you could do is derive a general solution of the individual convolutions of the form
$$u[n-n_0]\star u[n-n_1]=(n-N_0)u[n-N_1]\tag{1}$$
I leave it up to you to derive the correct values of $N_0$ and $N_1$ in terms of $n_0$ and $n_1$. Once you have obtained that result, you can just compute the solution to the original problem as the sum of four terms of the form $(1)$.