Integration of funtion multiplied by shifted heavyside step

[SpaceDomain]

As I understand it, the following is true:

<br /> \int_{0}^{\infty}{u(t - \lambda) d\lambda} = <br /> \int_{0}^{t}{d\lambda}<br />


But I do not understand why. It seems to me that the left side above should equal

<br /> \int_{\lambda}^{\infty}{d\lambda}<br />

since

<br /> u(t - \lambda) =<br /> \left\{\begin{array}{cc}0,&amp;\mbox{ if }<br /> t&lt; \lambda \\ 1, &amp; \mbox{ if } t&gt; \lambda \end{array}\right.<br />

I obviously don't understand this correctly. What am I not doing right?

 

[SpaceDomain]

I guess there is not a function multiplied by the unit step function here. But if there were, like:
<br /> <br /> \int_{0}^{\infty}{f( \lambda ) u(t - \lambda) d\lambda} <br /> <br />

then it would be equal to

<br /> \int_{0}^{t}{f( \lambda) d\lambda}<br />

Right?

 

[SpaceDomain]

Oh, I think I am seeing my mistake in logic here.

The unit step u(t - \lambda) is a shift the unit step u(\lambda) and then symmetric about the vertical axis.

I should be thinking of this as a graph of \lambda vs. u(\lambda).

I was getting thrown off by the variable t here because I am so used to it being the independent variable. Okay, okay. Never mind.