Simple yet tricky definite integration problem

[Tamis]

I'm given:
f(x)=\left\{\begin{matrix}<br /> 1 &amp; -1\leq x \leq 1\\ <br /> 0 &amp; otherwise <br /> \end{matrix}\right.

The integral to evaluate is:
\int_{-1}^1 f(x-t) dt

What integration techniques should i use to solve this problem?
Could someone please provide the steps to solve this problem (as the answer only provides the solution, and not the steps)?

 

[drawar]

I'm not sure if it really works, but let me just write it down:

Do the change of variable u=x-t then the integral becomes \int\limits_{x - 1}^{x + 1} {f(u)du}.
Now consider 4 cases:
(i) x - 1 \le - 1 &lt; 1 \le x + 1
(ii)- 1 \le x - 1 &lt; 1 \le x + 1
(iii)x - 1 \le - 1 &lt; x + 1 \le 1
(iv)- 1 \le x - 1 &lt; x + 1 \le 1

 

[Ray Vickson]

I'm given:
f(x)=\left\{\begin{matrix}<br /> 1 &amp; -1\leq x \leq 1\\ <br /> 0 &amp; otherwise <br /> \end{matrix}\right.

The integral to evaluate is:
\int_{-1}^1 f(x-t) dt

What integration techniques should i use to solve this problem?
Could someone please provide the steps to solve this problem (as the answer only provides the solution, and not the steps)?

Forum rules require you to show your work. Surely you can figure out most of this problem by yourself! Just think about what f(x-t) is for various x and t.