Integration with steps $\int_{-\infty}^\infty f(t-u) \, du $

Question

How do I solve this with steps? $$\int_{-\infty}^\infty f(t-u) \, du $$

where $$f(t)= \begin{cases} 0& \text{if}\quad t<0\\ 1 & \text{if}\quad 0\leq t\leq1\\ 0 & \text{if}\quad t>1. \end{cases} $$

Answer 1

enforcing the change of variables $x=t-u\implies du=-dx$ then $$\int_{-\infty}^\infty f(t-u) \, du =\int_{-\infty}^\infty f(x)dx= \int_{0}^1 \, dx=1$$

Alternatively $$0\le t-u\le1 \Longleftrightarrow t-1\le u \le t$$

Hence $$\int_{-\infty}^\infty f(t-u) \, du =\int_{-\infty}^\infty 1_{[0,1]}(t-u) \, du= \int_{t-1}^t \, du =1$$