- #1

fred2028

- 19

- 0

I know that

[tex]\[\int {u(t-a)dt = tu(t-a)} \][/tex]

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,

[tex]\[\int {u(t-a)\cos tdt = u(t-a)\int {\cos tdt} } \][/tex]

and not

[tex]\[\int {u(t-a)\cos tdt = u(t-a)\sin t - \int {tu(t-a)\cos tdt} } \][/tex]

?

And my 2nd question is how in the world do you solve

[tex]\[\int {u( - t + a)dt} \][/tex]

and how would you graph this?

Nextly, I assume that

[tex]\[f(t)u(t - a)u(t - b) = f(t),t \geqslant a\][/tex]

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?

[tex]\[f(t)u(t - 1)u(t + 1)\][/tex]