1. ## Shifted fcn...

Given:

x(t) = u(t) - u(t-10)

calculate:

Int(x^2)tdt

Am I thinking about this correctly... u(t) is 0 from neg. infinity to 0, and 1 after. u(t-10) is the same thing just shifted to the right to "turn on" at 10. So in effect we have a rectangle of amplitude 1 with bounds from 0 to 10. So our integral becomes:

int (1)^2 dt (with bounds of 0 and 10), which = 10?

2. your thinking won't be true in general (and we don't even have a definite integral here). Take $u(t) = \sin t$, for instance. I think we need more info here, if you have some definite answer to get to, as opposed to an expression.

3. I'm assuming that $u(t)$ is intended to notate the Heaviside step function, not an arbitrary function.

In any case, could you clarify whether this is a definite or indefinite integral? If we take it as written, it would be an indefinite integral, i.e.,

$\int x^2(t) \, dt$

As you observed, you could define $x^2(t)$ as a piecewise function like so:

$x^2(t) = \begin{cases} 0, & \mbox{if } t<0 \\ 1, & \mbox{if } 0 \le t < 10 \\ 0, & \mbox{if } t \ge 10 \end{cases}$

Then we could find the anti-derivative using piecewise techniques.

If instead you are calculating a definite integral whose bounds cover the 0 to 10 range, then your work is correct.