I know that
\(\displaystyle \[\int {u(t-a)dt = tu(t-a)} \]\)
where u(t) is the unit step function. Does this mean that any arbitrary function multiplied by u(t) is just the integral of that function multiplied by u(t), and I don't have to do integration by parts? For example,
\(\displaystyle \[\int {u(t-a)\cos tdt = u(t-a)\int {\cos tdt} } \]\)
and not
\(\displaystyle \[\int {u(t-a)\cos tdt = u(t-a)\sin t - \int {tu(t-a)\cos tdt} } \]\)
?
And my 2nd question is how in the world do you solve
\(\displaystyle \[\int {u( - t + a)dt} \]\)
and how would you graph this?
Nextly, I assume that
\(\displaystyle \[f(t)u(t - a)u(t - b) = f(t),t \geqslant a\]\)
if a > b. Basically, the function only "turns on" when both unit steps are 1. Is this true? If so, would the following equal 0 since the 2 unit step functions do not overlap?
\(\displaystyle \[f(t)u(t - 1)u(t + 1)\]\)
\(\displaystyle \[\int {u(t-a)dt = tu(t-a)} \]\)
where u(t) is the unit step function. Does this mean that any arbitrary function multiplied by u(t) is just the integral of that function multiplied by u(t), and I don't have to do integration by parts? For example,
\(\displaystyle \[\int {u(t-a)\cos tdt = u(t-a)\int {\cos tdt} } \]\)
and not
\(\displaystyle \[\int {u(t-a)\cos tdt = u(t-a)\sin t - \int {tu(t-a)\cos tdt} } \]\)
?
And my 2nd question is how in the world do you solve
\(\displaystyle \[\int {u( - t + a)dt} \]\)
and how would you graph this?
Nextly, I assume that
\(\displaystyle \[f(t)u(t - a)u(t - b) = f(t),t \geqslant a\]\)
if a > b. Basically, the function only "turns on" when both unit steps are 1. Is this true? If so, would the following equal 0 since the 2 unit step functions do not overlap?
\(\displaystyle \[f(t)u(t - 1)u(t + 1)\]\)