Is x(t) equal to the shifted step function u(t+2)-u(t)?

[Drao92]

When you shift step function u(t-a), is it true for a<0?
Im sorry but we did only exercises when you don't have to use this and in my homework using this is the only way i can solve it.
Like in this pic. Isnt x(t) (forgot to mark the axis, sorry) x(t)=u(t+2)-u(t)?
https://www.physicsforums.com/attachment.php?attachmentid=57584&stc=1&d=1365341968

 

[CompuChip]

The basic definition is
u(t) = \begin{cases} 0 &amp; t &lt; 0 \\ 1 &amp; t &gt; 0 \end{cases}
(and usually it's defined as u(0) = 1/2, but let's ignore that for now).

So the shifted step function is 1 only if t - a > 0, which means if t > a.

From your graph I gather that you want a function which is 1 if -2 < x < 0 and 0 otherwise. You can easily verify your answer x(t) = u(t + 2) - u(t) by considering the regions x < -2, -2 < x < 0 and x > 0 separately.
E.g. what do x(-3), x(-1) and x(1) work out to?