Express the functions in terms of unit step function It seems that I am on a conflict, because I have read few things, but it seems that I am rather confused now.
We have these two functions: 
$$
x(t)=
\begin{cases} 1, \quad 0\leq t\leq 1T, \\
0,\quad \text{else}.
\end{cases}  
$$
and 
$$
h(t)=
\begin{cases}
t, \quad 0\leq t\leq 2T,\\
0,\quad\text{else}.
\end{cases}
$$
My initial thought for the answer was this: 
$x(t)=u(t)$ and for $h(t)=u(t)t$. I am sure that I am definitely doing something wrong here.  Thanks for reading my question :)
p.s. I had to re upload the question due a error. Cheers
 A: I, along with Gautam, would also say that drawing pictures is the best way of going about this. If you are unable or unwilling to draw pictures, however, one alternative recourse would be to use Iverson brackets, and then convert to unit step functions afterward.
Recall that the Iverson bracket $[p]$ is $1$ if the condition $p$ is true, and $0$ if $p$ is false. With this, it is easy to rewrite a piecewise function in terms of Iverson brackets. Using $h(t)$ as an example, we have the Iversonian form
$$h(t)=t[0\leq t\leq2T]$$
The Iverson bracket possesses the very convenient identity $[p\text{ and }q]=[p][q]$; applying this to your function yields
$$h(t)=t[0\leq t][t\leq2T]=t[t\geq0][2T-t\geq0]$$
Now, another convenient identity exists between the unit step function $\mathfrak u(t)$ and the Iverson bracket: $\mathfrak u(t)=[t\geq 0]$; we thus have
$$h(t)=t\mathfrak u(t)\mathfrak u(2T-t)$$
I'll also note that $\mathfrak u(t)\mathfrak u(2T-t)$ and $\mathfrak u(t)-\mathfrak u(t-2T)$ have very different behavior at $t=2T$; the former is equal to $1$, while the latter is equal to $0$ at $t=2T$. The latter form does not square with the original piecewise definition that the function be equal to the identity function within the closed interval $[0,2T]$.
One can of course do a similar derivation for $x(t)$; this is left as an exercise for the interested reader.
A: The first one is
$$ x(t) = u(t) - u(t-T)$$
The second one is 
$$ h(t) = \int_{0}^{t} u(x)dx \quad 0\leq t \leq 2T$$
       $$ = t[u(t) - u(t-2T)]$$
