Integral of a function multiplied by a Heaviside step functi

[fred2028]

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 $$f(t)u(t - 1)u(t + 1)$$\)